From 60231fd08aac86615475561e2b5a6662c9e98149 Mon Sep 17 00:00:00 2001 From: jemmybutton Date: Thu, 30 May 2024 22:46:20 +0300 Subject: [PATCH] migrated from cleveref to zref-clever --- byrne-en-latex.tex | 1386 ++++++++++++++++++++--------------------- byrne-ru-latex.tex | 1475 ++++++++++++++++++++++---------------------- byrnebook.cls | 304 ++++++++- 3 files changed, 1713 insertions(+), 1452 deletions(-) diff --git a/byrne-en-latex.tex b/byrne-en-latex.tex index 8441889..95f40b2 100644 --- a/byrne-en-latex.tex +++ b/byrne-en-latex.tex @@ -1,4 +1,4 @@ -\documentclass{byrnebook} +\documentclass[booklanguage=english]{byrnebook} %\usepackage{lua-visual-debug} \begin{document} @@ -333,12 +333,12 @@ \part*{Introduction} we have $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$,\\ \drawAngle{BAC} common and $\drawUnitLine{AB} = \drawUnitLine{AC}$:\\ $\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$\\ -and $\drawAngle{CEB} = \drawAngle{BDC}$ \bycref{prop:I.IV}.\\ +and $\drawAngle{CEB} = \drawAngle{BDC}$ \byref{prop:I.IV}.\\ Again in \drawLine{BC,BE,CE} and \drawLine{BC,BD,CD},\\ $\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{CEB} = \drawAngle{BDC}$\\ and $\drawUnitLine{BE} = \drawUnitLine{CD}$;\\ $\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ -and $\drawAngle{DCB} = \drawAngle{CBE}$ \bycref{prop:I.IV}\\ +and $\drawAngle{DCB} = \drawAngle{CBE}$ \byref{prop:I.IV}\\ But $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$. \end{center} @@ -348,9 +348,9 @@ \part*{Introduction} \emph{By annexing Letters to the Diagram.} \end{center} -Let the equal sides AB and AC be produced through the extremities BC, of the third side, and in the produced part BD of either, let any point D be assumed, and from the other let AE be cut off equal to AD \bycref{prop:I.III}. Let points E and D, so taken in the produced sides, be connected by straight lines DC and BE with the alternate extremities of the third side of the triangle. +Let the equal sides AB and AC be produced through the extremities BC, of the third side, and in the produced part BD of either, let any point D be assumed, and from the other let AE be cut off equal to AD \byref{prop:I.III}. Let points E and D, so taken in the produced sides, be connected by straight lines DC and BE with the alternate extremities of the third side of the triangle. -In the triangles DAC and EAB the sides DA and AC are respectively equal to EA and AB, and the included angle A is common to both triangles. Hence \bycref{prop:I.IV} the line DC is equal to BE, the angle ADC to the angle AEB, and the angle ACD to the angle ABE; if from the equal lines AD and AE the equal sides AB and AC be taken, the remainders BD and CE will be equal. Hence in the triangles BDC and CEB, the sides BD and DC are respectively equal to CE and EB, and the angles D and E included by those sides are also equal. Hence \bycref{prop:I.IV} the angles DBC and ECB, which are those included by the third side BC and the productions of the equal sides AB and AC are equal. Also the angles DCB and EBC are equal if those equals be taken from the angles DCA and EBA before proved equal, the remainders, which are the angles ABC and ACB opposite to the equal sides, will be equal. +In the triangles DAC and EAB the sides DA and AC are respectively equal to EA and AB, and the included angle A is common to both triangles. Hence \byref{prop:I.IV} the line DC is equal to BE, the angle ADC to the angle AEB, and the angle ACD to the angle ABE; if from the equal lines AD and AE the equal sides AB and AC be taken, the remainders BD and CE will be equal. Hence in the triangles BDC and CEB, the sides BD and DC are respectively equal to CE and EB, and the angles D and E included by those sides are also equal. Hence \byref{prop:I.IV} the angles DBC and ECB, which are those included by the third side BC and the productions of the equal sides AB and AC are equal. Also the angles DCB and EBC are equal if those equals be taken from the angles DCA and EBA before proved equal, the remainders, which are the angles ABC and ACB opposite to the equal sides, will be equal. \emph{Therefore in an isosceles triangle,} \&c. @@ -1091,13 +1091,13 @@ \chapter*{Propositions} draw byLabelLineEnd(A, B, 1); draw byLabelLineEnd(B, A, 0); }} -\bycref{post:I.III};\\ -draw \drawUnitLine{CA} and \drawUnitLine{BC} \bycref{post:I.I}.\\ +\byref{post:I.III};\\ +draw \drawUnitLine{CA} and \drawUnitLine{BC} \byref{post:I.I}.\\ Then will \drawLine[bottom][triangleABC]{AB,CA,BC} be equilateral. -For $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{def:I.XV};\\ -and $\drawUnitLine{AB} = \drawUnitLine{BC}$ \bycref{def:I.XV};\\ -$\therefore \drawUnitLine{CA} = \drawUnitLine{BC}$ \bycref{ax:I.I};\\ +For $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{def:I.XV};\\ +and $\drawUnitLine{AB} = \drawUnitLine{BC}$ \byref{def:I.XV};\\ +$\therefore \drawUnitLine{CA} = \drawUnitLine{BC}$ \byref{ax:I.I};\\ and therefore \triangleABC\ is the equilateral triangle required. \end{center} @@ -1143,7 +1143,7 @@ \chapter*{Propositions} }), to draw a straight line equal to a given straight line (\drawUnitLine{BC}).} \begin{center} -Draw \drawUnitLine{AB} \bycref{post:I.I}, describe \drawFromCurrentPicture[bottom]{ +Draw \drawUnitLine{AB} \byref{post:I.I}, describe \drawFromCurrentPicture[bottom]{ startAutoLabeling; startTempScale(scaleFactor*3); startGlobalRotation(180-lineAngle.AB); @@ -1151,8 +1151,8 @@ \chapter*{Propositions} stopGlobalRotation; stopTempScale; stopAutoLabeling; -} \bycref{prop:I.I},\\ -produce \drawUnitLine{BD} \bycref{post:I.II},\\ +} \byref{prop:I.I},\\ +produce \drawUnitLine{BD} \byref{post:I.II},\\ describe \drawFromCurrentPicture{ draw byNamedLine (BC); @@ -1160,21 +1160,21 @@ \chapter*{Propositions} draw byLabelLineEnd(B, C, 0); draw byLabelLineEnd(C, B, 0); } -\bycref{post:I.III}, and +\byref{post:I.III}, and \drawFromCurrentPicture{ draw byNamedLineSeq(0)(BD, BE); draw byNamedCircle(A); draw byLabelLineEnd(D, E, 0); draw byLabelLineEnd(E, D, 1); } -\bycref{post:I.III};\\ -produce \drawUnitLine{DA} \bycref{post:I.II},\\ +\byref{post:I.III};\\ +produce \drawUnitLine{DA} \byref{post:I.II},\\ then \drawUnitLine{AF} is the line required. -For $\drawUnitLine{BE,BD} = \drawUnitLine{DA,AF}$ \bycref{def:I.XV},\\ -and $\drawUnitLine{BD} = \drawUnitLine{DA}$ \bycref{\constref},\\ -$\therefore \drawUnitLine{BE} = \drawUnitLine{AF}$ \bycref{post:I.III},\\ -but \bycref{def:I.XV} $\drawUnitLine{BC} = \drawUnitLine{BE} = \drawUnitLine{AF}$; +For $\drawUnitLine{BE,BD} = \drawUnitLine{DA,AF}$ \byref{def:I.XV},\\ +and $\drawUnitLine{BD} = \drawUnitLine{DA}$ \byref{\constref},\\ +$\therefore \drawUnitLine{BE} = \drawUnitLine{AF}$ \byref{post:I.III},\\ +but \byref{def:I.XV} $\drawUnitLine{BC} = \drawUnitLine{BE} = \drawUnitLine{AF}$; $\therefore \drawUnitLine{AF}$ drawn from the given point (\pointA), is equal to the given line \drawUnitLine{BC}. \end{center} @@ -1209,20 +1209,20 @@ \chapter*{Propositions} \problem{F}{rom}{the greater (\drawUnitLine{AB,BC}) of two given straight lines, to cut off a part equal to the less (\drawUnitLine{EF}).} \begin{center} -Draw $\drawUnitLine{AD} = \drawUnitLine{EF}$ \bycref{prop:I.II};\\ +Draw $\drawUnitLine{AD} = \drawUnitLine{EF}$ \byref{prop:I.II};\\ describe \drawFromCurrentPicture{ draw byNamedLine (AD); draw byNamedCircle(A); draw byLabelLineEnd(D, A, 0); draw byLabelLineEnd(A, D, 0); -} \bycref{post:I.III},\\ +} \byref{post:I.III},\\ then $\drawUnitLine{EF} = \drawUnitLine{AB}$ -For $\drawUnitLine{AD} = \drawUnitLine{AB}$ \bycref{def:I.XV},\\ -and $\drawUnitLine{EF} = \drawUnitLine{AD}$ \bycref{\constref}; +For $\drawUnitLine{AD} = \drawUnitLine{AB}$ \byref{def:I.XV},\\ +and $\drawUnitLine{EF} = \drawUnitLine{AD}$ \byref{\constref}; -$\therefore \drawUnitLine{EF} = \drawUnitLine{AB}$ \bycref{ax:I.I}. +$\therefore \drawUnitLine{EF} = \drawUnitLine{AB}$ \byref{ax:I.I}. \end{center} \qed @@ -1259,7 +1259,7 @@ \chapter*{Propositions} \drawCurrentPictureInMargin \problem[4]{I}{f}{two triangles have two sides of the one respectively equal to two sides of the other, (\drawUnitLine{AB} to \drawUnitLine{DE} and \drawUnitLine{CA} to \drawUnitLine{FD}) and the angles (\drawAngle{A} and \drawAngle{D}) contained by those equal sides also equal; then their bases or their sides (\drawUnitLine{BC} and \drawUnitLine{EF}) are also equal: and the remaining angles opposite to equal sides are respectively equal ($\drawAngle{B} = \drawAngle{E}$ and $\drawAngle{C} = \drawAngle{F}$): and the triangles are equal in every respect.} -Let two triangles be conceived, to be so placed, that the vertex of the one of the equal angles, \drawAngle{A} or \drawAngle{D}; shall fall upon that of the other, and \drawUnitLine{AB} to coincide with \drawUnitLine{DE}, then will \drawUnitLine{CA} coincide with \drawUnitLine{FD} if applied: consequently \drawUnitLine{BC} will coincide with \drawUnitLine{EF}, or two straight lines will enclose a space, which is impossible \bycref{ax:I.X}, therefore $\drawUnitLine{BC} = \drawUnitLine{EF}$, $\drawAngle{B} = \drawAngle{E}$ and $\drawAngle{C} = \drawAngle{F}$, and as the triangles \drawLine{CA,BC,AB} and \drawLine{FD,EF,DE} coincide, when applied, they are equal in every respect. +Let two triangles be conceived, to be so placed, that the vertex of the one of the equal angles, \drawAngle{A} or \drawAngle{D}; shall fall upon that of the other, and \drawUnitLine{AB} to coincide with \drawUnitLine{DE}, then will \drawUnitLine{CA} coincide with \drawUnitLine{FD} if applied: consequently \drawUnitLine{BC} will coincide with \drawUnitLine{EF}, or two straight lines will enclose a space, which is impossible \byref{ax:I.X}, therefore $\drawUnitLine{BC} = \drawUnitLine{EF}$, $\drawAngle{B} = \drawAngle{E}$ and $\drawAngle{C} = \drawAngle{F}$, and as the triangles \drawLine{CA,BC,AB} and \drawLine{FD,EF,DE} coincide, when applied, they are equal in every respect. \qed @@ -1295,8 +1295,8 @@ \chapter*{Propositions} \problem[4]{I}{n}{any isosceles triangle \drawLine[bottom]{BC,AC,AB} if the equal sides be produced, the external angles at the base are equal, and the internal angles at the base are also equal.} \begin{center} -Produce \drawUnitLine{AB} and \drawUnitLine{AC} \bycref{post:I.II},\\ -take $\drawUnitLine{BD} = \drawUnitLine{CE}$ \bycref{prop:I.III};\\ +Produce \drawUnitLine{AB} and \drawUnitLine{AC} \byref{post:I.II},\\ +take $\drawUnitLine{BD} = \drawUnitLine{CE}$ \byref{prop:I.III};\\ draw \drawUnitLine{BE} and \drawUnitLine{CD}. Then in @@ -1313,15 +1313,15 @@ \chapter*{Propositions} draw byNamedLineSeq(0)(BD,CD,AC,AB); stopAutoLabeling; }\\ -we have $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$ \bycref{\constref},\\ +we have $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$ \byref{\constref},\\ \drawAngle{BAC} common to both,\\ -and $\drawUnitLine{AB} = \drawUnitLine{AC}$ \bycref{\hypref}\\ -$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ and $\drawAngle{CEB} = \drawAngle{BDC}$ \bycref{prop:I.IV}. +and $\drawUnitLine{AB} = \drawUnitLine{AC}$ \byref{\hypref}\\ +$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ and $\drawAngle{CEB} = \drawAngle{BDC}$ \byref{prop:I.IV}. Again in \drawLine{BE,CE,BC} and \drawLine{BD,CD,BC}\\ we have $\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{CEB} = \drawAngle{BDC}$ and $\drawUnitLine{BE} = \drawUnitLine{CD}$,\\ -$\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ and $\drawAngle{DCB} = \drawAngle{CBE}$ \bycref{prop:I.IV}\\ -but $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$ \bycref{ax:I.III}. +$\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ and $\drawAngle{DCB} = \drawAngle{CBE}$ \byref{prop:I.IV}\\ +but $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$ \byref{ax:I.III}. \end{center} \qed @@ -1348,14 +1348,14 @@ \chapter*{Propositions} \drawCurrentPictureInMargin \problem{I}{n}{any triangle (\drawLine[bottom][triangleABD]{CA,CD,BD,AB}) if two angles (\drawAngle{A} and \drawAngle{B}) are equal, the sides (\drawUnitLine{CA,CD} and \drawUnitLine{BD}) opposite to them are also equal.} -For if the sides be not equal, let one of them \drawUnitLine{CA,CD} be greater than the other \drawUnitLine{BD}, and from it to cut off $\drawUnitLine{CA} = \drawUnitLine{BD}$ \bycref{prop:I.III}, draw \drawUnitLine{BC}. +For if the sides be not equal, let one of them \drawUnitLine{CA,CD} be greater than the other \drawUnitLine{BD}, and from it to cut off $\drawUnitLine{CA} = \drawUnitLine{BD}$ \byref{prop:I.III}, draw \drawUnitLine{BC}. \begin{center} Then in \drawLine[bottom]{BC,AB,CA} and \triangleABD,\\ -$\drawUnitLine{CA} = \drawUnitLine{BD}$ \bycref{\constref},\\ -$\drawAngle{A} = \drawAngle{B}$ \bycref{\hypref}\\ +$\drawUnitLine{CA} = \drawUnitLine{BD}$ \byref{\constref},\\ +$\drawAngle{A} = \drawAngle{B}$ \byref{\hypref}\\ and \drawUnitLine{AB} common,\\ -$\therefore$ the triangles are equal \bycref{prop:I.IV}\\ +$\therefore$ the triangles are equal \byref{prop:I.IV}\\ a part equal to the whole, which is absurd;\\ $\therefore$ neither of the sides \drawUnitLine{CA,CD} or \drawUnitLine{BD} is greater than the other,\\ hence they are equal. @@ -1420,14 +1420,14 @@ \chapter*{Propositions} , then draw \drawUnitLine{CD} and, \begin{center} -$\drawAngle{C,DCB} = \drawAngle{CDA}$ \bycref{prop:I.V} +$\drawAngle{C,DCB} = \drawAngle{CDA}$ \byref{prop:I.V} $\therefore \drawAngle{DCB} < \drawAngle{CDA}$ and $\left. \begin{aligned} \therefore\drawAngle{DCB} &< \drawAngle{CDA,D}\\ - \mbox{but \bycref{prop:I.V}} \drawAngle{DCB} &= \drawAngle{CDA,D}\\ + \mbox{but \byref{prop:I.V}} \drawAngle{DCB} &= \drawAngle{CDA,D}\\ \end{aligned} \right\}\mbox{which is absurd,}$ \end{center} @@ -1512,16 +1512,16 @@ \chapter*{Propositions} \problem{T}{o}{bisect a given rectilinear angle (\drawAngle{BAD,CAD}).} \begin{center} -Take $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{prop:I.III} +Take $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{prop:I.III} -draw \drawUnitLine{BC}, upon which describe \drawLine{CD,DB,BC} \bycref{prop:I.I},\\ +draw \drawUnitLine{BC}, upon which describe \drawLine{CD,DB,BC} \byref{prop:I.I},\\ draw \drawUnitLine{AD}. -$\because \drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{\constref}\\ %Because +$\because \drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{\constref}\\ %Because and \drawUnitLine{AD} common to the two triangles\\ % what triangles? -and $\drawUnitLine{CD} = \drawUnitLine{DB}$ \bycref{\constref}, +and $\drawUnitLine{CD} = \drawUnitLine{DB}$ \byref{\constref}, -$\therefore \drawAngle{BAD} = \drawAngle{CAD}$ \bycref{prop:I.VIII}. +$\therefore \drawAngle{BAD} = \drawAngle{CAD}$ \byref{prop:I.VIII}. \end{center} \qed @@ -1548,12 +1548,12 @@ \chapter*{Propositions} \problem{T}{o}{bisect a given finite straight line (\drawUnitLine{DB,CD}).} \begin{center} -Construct \drawLine[bottom]{AB,CA,CD,DB} \bycref{prop:I.I},\\ -draw \drawUnitLine{AD}, making $\drawAngle{BAD} = \drawAngle{CAD}$ \bycref{prop:I.IX}. +Construct \drawLine[bottom]{AB,CA,CD,DB} \byref{prop:I.I},\\ +draw \drawUnitLine{AD}, making $\drawAngle{BAD} = \drawAngle{CAD}$ \byref{prop:I.IX}. -Then $\drawUnitLine{BD} = \drawUnitLine{DC}$ by \bycref{prop:I.IV}, +Then $\drawUnitLine{BD} = \drawUnitLine{DC}$ by \byref{prop:I.IV}, -for~$\drawUnitLine{AB} = \drawUnitLine{AC}$ \bycref{\constref} $\drawAngle{BAD} = \drawAngle{CAD}$\\ +for~$\drawUnitLine{AB} = \drawUnitLine{AC}$ \byref{\constref} $\drawAngle{BAD} = \drawAngle{CAD}$\\ and \drawUnitLine{AD} common to the two triangles. % what triangles? obviously DBA and DAC Therefore the given line is bisected. @@ -1590,17 +1590,17 @@ \chapter*{Propositions} \begin{center} Take any point (\drawPointL[middle][CA]{C}) in the given line,\\ -cut off $\drawUnitLine{DB} = \drawUnitLine{CD}$ \bycref{prop:I.III},\\ -construct \drawLine[bottom]{AB,CA,CD,DB} \bycref{prop:I.I},\\ +cut off $\drawUnitLine{DB} = \drawUnitLine{CD}$ \byref{prop:I.III},\\ +construct \drawLine[bottom]{AB,CA,CD,DB} \byref{prop:I.I},\\ draw \drawUnitLine{AD} and it shall be perpendicular to the given line. -For $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{\constref}\\ -$\drawUnitLine{CD} = \drawUnitLine{DB}$ \bycref{\constref}\\ +For $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{\constref}\\ +$\drawUnitLine{CD} = \drawUnitLine{DB}$ \byref{\constref}\\ and \drawUnitLine{AD} common to the two triangles. % what triangles? obviously DBA and DAC -Therefore $\drawAngle{ADB} = \drawAngle{CDA}$ \bycref{prop:I.VIII} +Therefore $\drawAngle{ADB} = \drawAngle{CDA}$ \byref{prop:I.VIII} -$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \bycref{def:I.X}. +$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \byref{def:I.X}. \end{center} \qed @@ -1646,17 +1646,17 @@ \chapter*{Propositions} \begin{center} With the given point \drawPointL{A} as centre, at one side of the line, and any distance \drawUnitLine{DB} capable of extending to the other side, describe \drawArc{O}. % improvement: the distance mentioned here seems not to be in the original drawing, it should either be drawn, or not be referenced graphically as DB -Make $\drawUnitLine{DB} = \drawUnitLine{CD}$ \bycref{prop:I.X},\\ +Make $\drawUnitLine{DB} = \drawUnitLine{CD}$ \byref{prop:I.X},\\ draw \drawUnitLine{AB}, \drawUnitLine{CA} and \drawUnitLine{AD},\\ then $\drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$. -For \bycref{prop:I.VIII} since $\drawUnitLine{DB} = \drawUnitLine{CD}$ \bycref{\constref},\\ +For \byref{prop:I.VIII} since $\drawUnitLine{DB} = \drawUnitLine{CD}$ \byref{\constref},\\ \drawUnitLine{AD} common to both,\\ % to both what? obviously DBA and DAC -and $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{def:I.XV}, +and $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{def:I.XV}, $\therefore \drawAngle{ADB} = \drawAngle{CDA}$, and -$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \bycref{def:I.X}. +$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \byref{def:I.X}. \end{center} \qed @@ -1685,14 +1685,14 @@ \chapter*{Propositions} \begin{center} If \drawUnitLine{ED} be $\perp$ to \drawUnitLine{BC} then,\\ -\drawAngle{ADB,EDA} and $\drawAngle{CDE} = \drawTwoRightAngles$ \bycref{def:I.X}, % improvement: def. 7 in the original is likely to be a mistake +\drawAngle{ADB,EDA} and $\drawAngle{CDE} = \drawTwoRightAngles$ \byref{def:I.X}, % improvement: def. 7 in the original is likely to be a mistake but if \drawUnitLine{ED} be not $\perp$ to \drawUnitLine{BC},\\ -draw $\drawUnitLine{AD} \perp \drawUnitLine{BC}$ \bycref{prop:I.XI};\\ -$\drawAngle{ADB} +\drawAngle{CDE,EDA} = \drawTwoRightAngles$ \bycref{\constref},\\ +draw $\drawUnitLine{AD} \perp \drawUnitLine{BC}$ \byref{prop:I.XI};\\ +$\drawAngle{ADB} +\drawAngle{CDE,EDA} = \drawTwoRightAngles$ \byref{\constref},\\ $\drawAngle{ADB} = \drawAngle{CDE,EDA} = \drawAngle{EDA} + \drawAngle{CDE}$ -$\therefore \drawAngle{ADB} + \drawAngle{CDE,EDA} = \drawAngle{ADB} + \drawAngle{EDA} + \drawAngle{CDE}$ \bycref{ax:I.II} +$\therefore \drawAngle{ADB} + \drawAngle{CDE,EDA} = \drawAngle{ADB} + \drawAngle{EDA} + \drawAngle{CDE}$ \byref{ax:I.II} $= \drawAngle{ADB,EDA} + \drawAngle{CDE} = \drawTwoRightAngles$. \end{center} @@ -1729,7 +1729,7 @@ \chapter*{Propositions} but by the hypothesis $\drawAngle{BDA} + \drawAngle{CDA} = \drawTwoRightAngles$ -$\therefore\drawAngle{CDA,EDC} = \drawAngle{CDA}$ \bycref{ax:I.III}; which is absurd \bycref{ax:I.IX}. +$\therefore\drawAngle{CDA,EDC} = \drawAngle{CDA}$ \byref{ax:I.III}; which is absurd \byref{ax:I.IX}. $\therefore \drawUnitLine{ED}$ is not the continuation of \drawUnitLine{BD}, and the like may be demonstrated of any other straight line except \drawUnitLine{DC}, $\therefore \drawUnitLine{DC}$ is the continuation of \drawUnitLine{BD}. \end{center} @@ -1763,7 +1763,7 @@ \chapter*{Propositions} $\drawAngle{AED} + \drawAngle{CEA} = \drawTwoRightAngles$ % -$\therefore \drawAngle{BEC} = \drawAngle{AED}$. % And this is according to \bycref{ax:I.III} +$\therefore \drawAngle{BEC} = \drawAngle{AED}$. % And this is according to \byref{ax:I.III} In the same manner it may be shown that\\ $\drawAngle{CEA} = \drawAngle{DEB}$. @@ -1812,13 +1812,13 @@ \chapter*{Propositions} } \begin{center} -Make $\drawUnitLine{BE} = \drawUnitLine{EC}$ \bycref{prop:I.X};\\ +Make $\drawUnitLine{BE} = \drawUnitLine{EC}$ \byref{prop:I.X};\\ Draw \drawUnitLine{AE} and produce it until $\drawUnitLine{ED} = \drawUnitLine{AE}$;\\ draw \drawUnitLine{CD}. In \drawLine{BE,AE,AB} and \drawLine{EC,ED,CD};\\ -$\drawUnitLine{BE} = \drawUnitLine{EC}$, $\drawAngle{AEB} = \drawAngle{DEC}$ and $\drawUnitLine{AE} = \drawUnitLine{ED}$ \bycref{\constref,prop:I.XV},\\ -$\therefore \drawAngle{B} = \drawAngle{ECD}$ \bycref{prop:I.IV},\\ +$\drawUnitLine{BE} = \drawUnitLine{EC}$, $\drawAngle{AEB} = \drawAngle{DEC}$ and $\drawUnitLine{AE} = \drawUnitLine{ED}$ \byref{\constref,prop:I.XV},\\ +$\therefore \drawAngle{B} = \drawAngle{ECD}$ \byref{prop:I.IV},\\ $\therefore \anglesECDpDCF\ > \drawAngle{B}$. In like manner it can be shown, that if \drawUnitLine{BC} be produced, $\drawAngle{GCA} > \drawAngle{A}$ \\ @@ -1853,7 +1853,7 @@ \chapter*{Propositions} Produce \drawUnitLine{AC}, then will\\ $\drawAngle{ACB} + \drawAngle{BCD} = \drawTwoRightAngles$. -But $\drawAngle{BCD} > \drawAngle{A}$ \bycref{prop:I.XVI} +But $\drawAngle{BCD} > \drawAngle{A}$ \byref{prop:I.XVI} $\therefore \drawAngle{ACB} + \drawAngle{A} < \drawTwoRightAngles$, \end{center} @@ -1891,11 +1891,11 @@ \chapter*{Propositions} \problem[3]{I}{n}{any triangle \drawLine{DC,BC,BA,AD} if one side \drawUnitLine{AD,DC} be greater than another \drawUnitLine{BC}, the angle opposite to the greater side is greater than the angle opposite to the less. I.\ e.\ $\drawAngle{DBC,ABD} > \drawAngle{A}$.} \begin{center} -Make $\drawUnitLine{DC} = \drawUnitLine{BC}$ \bycref{prop:I.III}, draw \drawUnitLine{DB}. +Make $\drawUnitLine{DC} = \drawUnitLine{BC}$ \byref{prop:I.III}, draw \drawUnitLine{DB}. -Then will $\drawAngle{D} = \drawAngle{DBC}$ \bycref{prop:I.V}; +Then will $\drawAngle{D} = \drawAngle{DBC}$ \byref{prop:I.V}; -but $\drawAngle{D} > \drawAngle{A}$ \bycref{prop:I.XVI} +but $\drawAngle{D} > \drawAngle{A}$ \byref{prop:I.XVI} $\therefore \drawAngle{DBC} > \drawAngle{A}$ and much more\\ is $\drawAngle{DBC,ABD} > \drawAngle{A}$. @@ -1926,11 +1926,11 @@ \chapter*{Propositions} $\drawUnitLine{BC} =$ or $< \drawUnitLine{CA}$. If $\drawUnitLine{BC} = \drawUnitLine{CA}$ then\\ -$\drawAngle{A} = \drawAngle{B}$ \bycref{prop:I.V};\\ +$\drawAngle{A} = \drawAngle{B}$ \byref{prop:I.V};\\ which is contrary to the hypothesis. \drawUnitLine{BC} is not less than \drawUnitLine{CA}; for if it were,\\ -$\drawAngle{A} < \drawAngle{B}$ \bycref{prop:I.XVIII}\\ +$\drawAngle{A} < \drawAngle{B}$ \byref{prop:I.XVIII}\\ which is contrary to the hypothesis: $\therefore \drawUnitLine{BC} > \drawUnitLine{CA}$. @@ -1963,15 +1963,15 @@ \chapter*{Propositions} \begin{center} Produce \drawUnitLine{DA}, and\\ -make $\drawUnitLine{CD} = \drawUnitLine{BD}$ \bycref{prop:I.III};\\ +make $\drawUnitLine{CD} = \drawUnitLine{BD}$ \byref{prop:I.III};\\ draw \drawUnitLine{BC}. -Then $\because \drawUnitLine{CD} = \drawUnitLine{BD}$ \bycref{\constref},\\ %because -$\drawAngle{CBD} = \drawAngle{C}$ \bycref{prop:I.V} +Then $\because \drawUnitLine{CD} = \drawUnitLine{BD}$ \byref{\constref},\\ %because +$\drawAngle{CBD} = \drawAngle{C}$ \byref{prop:I.V} -$\therefore \drawAngle{CBD,DBA} > \drawAngle{C}$ \bycref{ax:I.IX} +$\therefore \drawAngle{CBD,DBA} > \drawAngle{C}$ \byref{ax:I.IX} -$\therefore \drawUnitLine{DA} + \drawUnitLine{CD} > \drawUnitLine{AB}$ \bycref{prop:I.XIX} +$\therefore \drawUnitLine{DA} + \drawUnitLine{CD} > \drawUnitLine{AB}$ \byref{prop:I.XIX} and $\therefore \drawUnitLine{DA} + \drawUnitLine{BD} > \drawUnitLine{AB}$. \end{center} @@ -2008,17 +2008,17 @@ \chapter*{Propositions} \begin{center} Produce \drawSizedLine{AD},\\ -$\drawSizedLine{CA} + \drawSizedLine{EC} > \drawSizedLine{AD,DE}$ \bycref{prop:I.XX},\\ +$\drawSizedLine{CA} + \drawSizedLine{EC} > \drawSizedLine{AD,DE}$ \byref{prop:I.XX},\\ add \drawSizedLine{BE} to each,\\ -$\drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD,DE} + \drawSizedLine{BE}$ \bycref{ax:I.IV} +$\drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD,DE} + \drawSizedLine{BE}$ \byref{ax:I.IV} in the same manner it may be shown that\\ $\drawSizedLine{AD,DE} + \drawSizedLine{BE} > \drawSizedLine{AD} + \drawSizedLine{BD}$,\\ $\therefore \drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD} + \drawSizedLine{BD}$,\\ which was to be proved. -Again $\drawAngle{E} > \drawAngle{C}$ \bycref{prop:I.XVI},\\ -and also $\drawAngle{D} > \drawAngle{E}$ \bycref{prop:I.XVI}, +Again $\drawAngle{E} > \drawAngle{C}$ \byref{prop:I.XVI},\\ +and also $\drawAngle{D} > \drawAngle{E}$ \byref{prop:I.XVI}, $\therefore \drawAngle{D} > \drawAngle{C}$. \end{center} @@ -2074,14 +2074,14 @@ \chapter*{Propositions} the sum of any two greater than the third, to construct a triangle whose sides shall be respectively equal to the given lines.} \begin{center} -Assume $\drawSizedLine{AB} = \drawSizedLine{L'}$ \bycref{prop:I.III}. +Assume $\drawSizedLine{AB} = \drawSizedLine{L'}$ \byref{prop:I.III}. $\left. \begin{aligned} \mbox{Draw } \drawSizedLine{BE} &= \drawSizedLine{L''}\\ \mbox{and } \drawSizedLine{AD} &= \drawSizedLine{L'''}\\ \end{aligned} - \right\}\mbox{\bycref{prop:I.II}.}$ + \right\}\mbox{\byref{prop:I.II}.}$ With \drawSizedLine{AD} and \drawSizedLine{BE} as radii, describe \drawFromCurrentPicture{ @@ -2093,7 +2093,7 @@ \chapter*{Propositions} draw byNamedLine(BE); draw byNamedCircle(B); draw byLabelLineEnd(B, E, 0); draw byLabelLineEnd(E, B, 0); -}} \bycref{post:I.III};\\ +}} \byref{post:I.III};\\ draw \drawSizedLine{CA} and \drawSizedLine{BC},\\ then will \drawLine[bottom]{CA,BC,AB} be the triangle required. @@ -2103,7 +2103,7 @@ \chapter*{Propositions} \drawSizedLine{BC} &= \drawSizedLine{BE} = \drawSizedLine{L''} \mbox{,} \\ \mbox{and } \drawSizedLine{CA} &= \drawSizedLine{AD} = \drawSizedLine{L'''} \mbox{.} \\ \end{aligned} - \right\}\mbox{\bycref{\constref}}$ + \right\}\mbox{\byref{\constref}}$ \end{center} \qed @@ -2144,11 +2144,11 @@ \chapter*{Propositions} Draw \drawUnitLine{BC} between any two points in the legs of the given angle. \begin{center} -Construct \drawLine[bottom]{HF,GH,FG} \bycref{prop:I.XXII}\\ +Construct \drawLine[bottom]{HF,GH,FG} \byref{prop:I.XXII}\\ so that $\drawUnitLine{FG} = \drawUnitLine{AB}$, $\drawUnitLine{HF} = \drawUnitLine{CA}$\\ and $\drawUnitLine{GH} = \drawUnitLine{BC}$. -Then $\drawAngle{A} = \drawAngle{F}$ \bycref{prop:I.VIII}. +Then $\drawAngle{A} = \drawAngle{F}$ \byref{prop:I.VIII}. \end{center} \qed @@ -2218,18 +2218,18 @@ \chapter*{Propositions} startAutoLabeling; draw byNamedAngleSides(BAC)(AB, CA); stopAutoLabeling; -} = \angleFEG$ \bycref{prop:I.XXIII},\\ -and $\drawUnitLine{CA} = \drawUnitLine{GE}$ \bycref{prop:I.III},\\ +} = \angleFEG$ \byref{prop:I.XXIII},\\ +and $\drawUnitLine{CA} = \drawUnitLine{GE}$ \byref{prop:I.III},\\ draw \drawUnitLine{CD} and \drawUnitLine{BC}. -$\because \drawUnitLine{CA} = \drawUnitLine{AD}$ \bycref{ax:I.I,\hypref,\constref}\\ %because -$\therefore \drawAngle{BDA,CDB} = \drawAngle{DCA}$ \bycref{prop:I.V}\\ +$\because \drawUnitLine{CA} = \drawUnitLine{AD}$ \byref{ax:I.I,\hypref,\constref}\\ %because +$\therefore \drawAngle{BDA,CDB} = \drawAngle{DCA}$ \byref{prop:I.V}\\ but $\drawAngle{CDB} < \drawAngle{DCA}$,\\ and $\therefore \drawAngle{CDB} < \drawAngle{DCA,ACB}$, -$\therefore \drawUnitLine{DB} > \drawUnitLine{BC}$ \bycref{prop:I.XIX} +$\therefore \drawUnitLine{DB} > \drawUnitLine{BC}$ \byref{prop:I.XIX} -but $\drawUnitLine{BC} = \drawUnitLine{FG}$ \bycref{prop:I.IV} +but $\drawUnitLine{BC} = \drawUnitLine{FG}$ \byref{prop:I.IV} $\therefore \drawUnitLine{DB} > \drawUnitLine{FG}$. \end{center} @@ -2267,12 +2267,12 @@ \chapter*{Propositions} $\drawAngle{A} =\mbox{, } > \mbox{ or } < \drawAngle{D}$ \drawAngle{A} is not equal to \drawAngle{D}\\ -for if $\drawAngle{A} = \drawAngle{D}$ then $\drawUnitLine{CB} = \drawUnitLine{FE}$ \bycref{prop:I.IV}\\ +for if $\drawAngle{A} = \drawAngle{D}$ then $\drawUnitLine{CB} = \drawUnitLine{FE}$ \byref{prop:I.IV}\\ which is contrary to the hypothesis; \drawAngle{A} is not less than \drawAngle{D}\\ for if $\drawAngle{A} < \drawAngle{D}$\\ -then $\drawUnitLine{CB} < \drawUnitLine{FE}$ \bycref{prop:I.XXIV},\\ +then $\drawUnitLine{CB} < \drawUnitLine{FE}$ \byref{prop:I.XXIV},\\ which is also contrary to the hypothesis: $\therefore \drawAngle{A} > \drawAngle{D}$. @@ -2327,13 +2327,13 @@ \chapter*{Propositions} \drawLine[bottom]{GD,EG,DE} we have\\ $\drawUnitLine{CA} = \drawUnitLine{GD}$, $\drawAngle{A} = \drawAngle{D}$, $\drawUnitLine{AB} = \drawUnitLine{DE}$;\\ $\therefore \drawAngle{B} = \drawAngle{GED}$ (pr. 4.)\\ -but $\drawAngle{B} = \drawAngle{GED,FEG}$ \bycref{\hypref} +but $\drawAngle{B} = \drawAngle{GED,FEG}$ \byref{\hypref} and therefore $\drawAngle{GED} = \drawAngle{GED,FEG}$, which is absurd;\\ hence neither of the sides \drawUnitLine{CA} and \drawUnitLine{GD,FG} is greater than the other;\\ and $\therefore$ they are equal; -$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$, and $\drawAngle{C} = \drawAngle{F}$, \bycref{prop:I.IV}. +$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$, and $\drawAngle{C} = \drawAngle{F}$, \byref{prop:I.IV}. \end{center} \vfill\pagebreak @@ -2376,12 +2376,12 @@ \chapter*{Propositions} Then in \drawLine[bottom]{CA,BC,AB} and \drawLine[bottom]{FD,FG,DG} we have $\drawUnitLine{CA} = \drawUnitLine{FD}$, $\drawUnitLine{AB} = \drawUnitLine{DG}$ and $\drawAngle{A} = \drawAngle{D}$, -$\therefore \drawAngle{B} = \drawAngle{G}$ \bycref{prop:I.IV}\\ -but $\drawAngle{B} = \drawAngle{E}$ \bycref{\hypref} +$\therefore \drawAngle{B} = \drawAngle{G}$ \byref{prop:I.IV}\\ +but $\drawAngle{B} = \drawAngle{E}$ \byref{\hypref} -$\therefore \drawAngle{G} = \drawAngle{E}$ which is absurd \bycref{prop:I.XVI}. +$\therefore \drawAngle{G} = \drawAngle{E}$ which is absurd \byref{prop:I.XVI}. -Consequently, neither of the sides \drawUnitLine{AB} or \drawUnitLine{DG,GE} is greater than the other, hence they must be equal. It follows (by \bycref{prop:I.IV}) that the triangles are equal in all respects. +Consequently, neither of the sides \drawUnitLine{AB} or \drawUnitLine{DG,GE} is greater than the other, hence they must be equal. It follows (by \byref{prop:I.IV}) that the triangles are equal in all respects. \end{center} \qed @@ -2419,7 +2419,7 @@ \chapter*{Propositions} If \drawUnitLine{CD} be not parallel to \drawUnitLine{AB} they shall meet when produced. -If it be possible, let those lines be not parallel, but meet when produced; then the external angle \drawAngle{HEB} is greater than \drawAngle{CFG} \bycref{prop:I.XVI}, but they are also equal \bycref{\hypref}, which is absurd: in the same manner it may be shown that they cannot meet on the other side; $\therefore$ they are parallel. +If it be possible, let those lines be not parallel, but meet when produced; then the external angle \drawAngle{HEB} is greater than \drawAngle{CFG} \byref{prop:I.XVI}, but they are also equal \byref{\hypref}, which is absurd: in the same manner it may be shown that they cannot meet on the other side; $\therefore$ they are parallel. \qed @@ -2453,16 +2453,16 @@ \chapter*{Propositions} \problem{I}{f}{a straight line (\drawUnitLine{GH}), cutting two other straight lines (\drawUnitLine{AB} and \drawUnitLine{CD}), makes the external equal to the internal and opposite angle, at the same side of the cutting line (namely $\drawAngle{GEA} = \drawAngle{CFG}$ or $\drawAngle{BEG} = \drawAngle{GFD}$), or if it makes the two internal angles at the same side (\drawAngle{GFD} and \drawAngle{HEB}, or \drawAngle{CFG} and \drawAngle{AEH}) together equal to two right angles, those two straight lines are parallel.} \begin{center} -First, if $\drawAngle{GEA} = \drawAngle{CFG}$, then $\drawAngle{GEA} = \drawAngle{HEB}$ \bycref{prop:I.XV},\\ -$\therefore \drawAngle{CFG} = \drawAngle{HEB} \therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXVII}. +First, if $\drawAngle{GEA} = \drawAngle{CFG}$, then $\drawAngle{GEA} = \drawAngle{HEB}$ \byref{prop:I.XV},\\ +$\therefore \drawAngle{CFG} = \drawAngle{HEB} \therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXVII}. Secondly, if $\drawAngle{CFG} + \drawAngle{AEH} = \drawTwoRightAngles$,\\ -then $\drawAngle{AEH} + \drawAngle{HEB} = \drawTwoRightAngles$ \bycref{prop:I.XIII},\\ -$\therefore \drawAngle{CFG} + \drawAngle{AEH} = \drawAngle{AEH} + \drawAngle{HEB}$ \bycref{ax:I.III} +then $\drawAngle{AEH} + \drawAngle{HEB} = \drawTwoRightAngles$ \byref{prop:I.XIII},\\ +$\therefore \drawAngle{CFG} + \drawAngle{AEH} = \drawAngle{AEH} + \drawAngle{HEB}$ \byref{ax:I.III} $\therefore \drawAngle{CFG} = \drawAngle{HEB}$ -$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXVII}. +$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXVII}. \end{center} \qed @@ -2502,11 +2502,11 @@ \chapter*{Propositions} \drawCurrentPictureInMargin \problem{A}{ straight}{line (\drawUnitLine{GH}) falling on two parallel straight lines (\drawUnitLine{AB} and \drawUnitLine{CD}), makes the alternate angles equal to one another; and also the external equal to the internal and opposite angle on the same side; and the two internal angles on the same side together equal to two right angles.} -For if the alternate angles \drawAngle{IEA,HEI} and \drawAngle{GFD} be not equal, draw \drawUnitLine{IE}, making $\drawAngle{HEI} = \drawAngle{GFD}$ \bycref{prop:I.XXIII}. +For if the alternate angles \drawAngle{IEA,HEI} and \drawAngle{GFD} be not equal, draw \drawUnitLine{IE}, making $\drawAngle{HEI} = \drawAngle{GFD}$ \byref{prop:I.XXIII}. -Therefore $\drawUnitLine{IE,EJ} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXVII} and therefore two straight lines which intersect are parallel to the same straight line, which is impossible \bycref{ax:I.XII}. +Therefore $\drawUnitLine{IE,EJ} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXVII} and therefore two straight lines which intersect are parallel to the same straight line, which is impossible \byref{ax:I.XII}. -Hence \drawAngle{IEA,HEI} and \drawAngle{GFD} are not unequal, that is, they are equal: $\drawAngle{IEA,HEI} = \drawAngle{GEB}$ \bycref{prop:I.XV}; $\therefore \drawAngle{GEB} = \drawAngle{GFD}$, the external angle equal to the internal and opposite on the same side: if \drawAngle{BEH} be added to both, then $\drawAngle{GFD} + \drawAngle{BEH} = \drawAngle{BEH,GEB} = \drawTwoRightAngles$ \bycref{prop:I.XIII}. That is to say, the two internal angles at the same side of the cutting line are equal to two right angles. +Hence \drawAngle{IEA,HEI} and \drawAngle{GFD} are not unequal, that is, they are equal: $\drawAngle{IEA,HEI} = \drawAngle{GEB}$ \byref{prop:I.XV}; $\therefore \drawAngle{GEB} = \drawAngle{GFD}$, the external angle equal to the internal and opposite on the same side: if \drawAngle{BEH} be added to both, then $\drawAngle{GFD} + \drawAngle{BEH} = \drawAngle{BEH,GEB} = \drawTwoRightAngles$ \byref{prop:I.XIII}. That is to say, the two internal angles at the same side of the cutting line are equal to two right angles. \qed @@ -2544,11 +2544,11 @@ \chapter*{Propositions} \begin{center} Let \drawUnitLine{JK} intersect $\left\{\vcenter{\nointerlineskip\hbox{\drawUnitLine{AB}}\nointerlineskip\hbox{\drawUnitLine{CD}}\nointerlineskip\hbox{\drawUnitLine{EF}}}\right\}$; -Then, $\drawAngle{G} = \drawAngle{I} = \drawAngle{H}$ \bycref{prop:I.XXIX}, +Then, $\drawAngle{G} = \drawAngle{I} = \drawAngle{H}$ \byref{prop:I.XXIX}, $\therefore \drawAngle{G} = \drawAngle{H}$ -$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{EF}$ \bycref{prop:I.XXVIII}. % improvement: byrne references I.XXVII here, which is wrong https://github.com/jemmybutton/byrne-euclid/issues/27#issuecomment-507012558 +$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{EF}$ \byref{prop:I.XXVIII}. % improvement: byrne references I.XXVII here, which is wrong https://github.com/jemmybutton/byrne-euclid/issues/27#issuecomment-507012558 \end{center} \qed @@ -2581,9 +2581,9 @@ \chapter*{Propositions} \begin{center} Draw \drawUnitLine{EF} from the point \drawPointL[middle][EB]{E} to any point \drawPointL[middle][CF]{F} in \drawUnitLine{CF,FD}, -make $\drawAngle{E} = \drawAngle{F}$ \bycref{prop:I.XXIII}, +make $\drawAngle{E} = \drawAngle{F}$ \byref{prop:I.XXIII}, -then $\drawUnitLine{AE,EB} \parallel \drawUnitLine{CF,FD}$ \bycref{prop:I.XXVII}. +then $\drawUnitLine{AE,EB} \parallel \drawUnitLine{CF,FD}$ \byref{prop:I.XXVII}. \end{center} \qed @@ -2615,19 +2615,19 @@ \chapter*{Propositions} \begin{center} Through the point \drawPointL[middle][AD,AE]{A} draw\\ -$\drawUnitLine{AE} \parallel \drawUnitLine{BC}$ \bycref{prop:I.XXXI}. +$\drawUnitLine{AE} \parallel \drawUnitLine{BC}$ \byref{prop:I.XXXI}. Then $\left\{ \begin{aligned} \drawAngle{DAE} &= \drawAngle{B}\\ \drawAngle{EAC} &= \drawAngle{C}\\ \end{aligned} - \right\}$ \bycref{prop:I.XXIX}, + \right\}$ \byref{prop:I.XXIX}, -$\therefore \drawAngle{B} + \drawAngle{C} = \drawAngle{DAE,EAC}$ \bycref{ax:I.II}, +$\therefore \drawAngle{B} + \drawAngle{C} = \drawAngle{DAE,EAC}$ \byref{ax:I.II}, and therefore\\ -$\drawAngle{B} + \drawAngle{CAB} + \drawAngle{C} = \drawAngle{DAE,EAC,CAB} = \drawTwoRightAngles$ \bycref{prop:I.XIII}. +$\drawAngle{B} + \drawAngle{CAB} + \drawAngle{C} = \drawAngle{DAE,EAC,CAB} = \drawTwoRightAngles$ \byref{prop:I.XIII}. \end{center} \qed @@ -2660,13 +2660,13 @@ \chapter*{Propositions} \begin{center} Draw \drawUnitLine{AD} the diagonal. -$\drawUnitLine{AB} = \drawUnitLine{CD}$ \bycref{\hypref}\\ -$\drawAngle{BAD} = \drawAngle{CDA}$ \bycref{prop:I.XXIX}\\ +$\drawUnitLine{AB} = \drawUnitLine{CD}$ \byref{\hypref}\\ +$\drawAngle{BAD} = \drawAngle{CDA}$ \byref{prop:I.XXIX}\\ and \drawUnitLine{AD} common to the two triangles; % what triangles? -$\therefore \drawUnitLine{AC} = \drawUnitLine{BD}$, and $\drawAngle{ADB} = \drawAngle{DAC}$ \bycref{prop:I.IV}; +$\therefore \drawUnitLine{AC} = \drawUnitLine{BD}$, and $\drawAngle{ADB} = \drawAngle{DAC}$ \byref{prop:I.IV}; -and $\therefore \drawUnitLine{AC} \parallel \drawUnitLine{BD}$ \bycref{prop:I.XXVII}. +and $\therefore \drawUnitLine{AC} \parallel \drawUnitLine{BD}$ \byref{prop:I.XXVII}. \end{center} \qed @@ -2704,7 +2704,7 @@ \chapter*{Propositions} \drawAngle{BAD} &= \drawAngle{CDA}\\ \drawAngle{DAC} &= \drawAngle{ADB}\\ \end{aligned} - \right\}$ \bycref{prop:I.XXIX} + \right\}$ \byref{prop:I.XXIX} and \drawUnitLine{AD} common to the two triangles. @@ -2714,11 +2714,11 @@ \chapter*{Propositions} \drawUnitLine{AC} &= \drawUnitLine{BD}\\ \drawAngle{B} &= \drawAngle{C}\\ \end{aligned} - \right\}$ \bycref{prop:I.XXVI}\\ -and $\drawAngle{BAD,DAC} = \drawAngle{CDA,ADB}$ \bycref{ax:I.II}. + \right\}$ \byref{prop:I.XXVI}\\ +and $\drawAngle{BAD,DAC} = \drawAngle{CDA,ADB}$ \byref{ax:I.II}. \end{center} -Therefore the opposite sides and angles of the parallelogram are equal: and as the triangles \drawLine{AD,CD,AC} and \drawLine{AB,BD,AD} are equal in every respect \bycref{prop:I.IV}, the diagonal divides the parallelogram into two equal parts. +Therefore the opposite sides and angles of the parallelogram are equal: and as the triangles \drawLine{AD,CD,AC} and \drawLine{AB,BD,AD} are equal in every respect \byref{prop:I.IV}, the diagonal divides the parallelogram into two equal parts. \qed @@ -2764,9 +2764,9 @@ \chapter*{Propositions} \end{aligned} \right\} \begin{aligned} - &\mbox{\bycref{prop:I.XXIX}}\\ - &\mbox{\bycref{prop:I.XXIX}}\\ - &\mbox{\bycref{prop:I.XXXIV}.}\\ + &\mbox{\byref{prop:I.XXIX}}\\ + &\mbox{\byref{prop:I.XXIX}}\\ + &\mbox{\byref{prop:I.XXXIV}.}\\ \end{aligned} $ @@ -2783,7 +2783,7 @@ \chapter*{Propositions} draw byNamedPolygon (EFDG, BEG); stopAutoLabeling; draw byNamedLine (BD); -}$ \bycref{prop:I.VIII} +}$ \byref{prop:I.VIII} $\therefore \drawFromCurrentPicture[middle][polygonAFDC]{ @@ -2839,19 +2839,19 @@ \chapter*{Propositions} \begin{center} Draw \drawUnitLine{CE} and \drawUnitLine{DF}\\ -$\drawUnitLine{CD} = \drawUnitLine{GH} = \drawUnitLine{EF}$ by \bycref{prop:I.XXXIV,\hypref}; +$\drawUnitLine{CD} = \drawUnitLine{GH} = \drawUnitLine{EF}$ by \byref{prop:I.XXXIV,\hypref}; $\therefore \drawUnitLine{CD} = \mbox{ and } \parallel \drawUnitLine{EF}$; -$\therefore \drawUnitLine{CE} = \mbox{ and } \parallel \drawUnitLine{DF}$ \bycref{prop:I.XXXIII} +$\therefore \drawUnitLine{CE} = \mbox{ and } \parallel \drawUnitLine{DF}$ \byref{prop:I.XXXIII} And therefore \drawPolygon[bottom][polygonCDFE]{CDI, IDJE, EFJ} is a parallelogram: -but $\polygonABDC = \polygonCDFE = \polygonEFHG$ \bycref{prop:I.XXXV} +but $\polygonABDC = \polygonCDFE = \polygonEFHG$ \byref{prop:I.XXXV} -$\therefore \polygonABDC = \polygonEFHG$ \bycref{ax:I.I}. +$\therefore \polygonABDC = \polygonEFHG$ \byref{ax:I.I}. \end{center} \qed @@ -2898,21 +2898,21 @@ \chapter*{Propositions} \mbox{Draw } \drawUnitLine{AC} &\parallel \drawUnitLine{BD}\\ \drawUnitLine{FD} &\parallel \drawUnitLine{EC}\\ \end{aligned} -\right\}\mbox{\bycref{prop:I.XXXI}}$ +\right\}\mbox{\byref{prop:I.XXXI}}$ Produce \drawUnitLine{HI}. \drawPolygon[bottom][polygonABDC]{ABC, BCG, CDG} and \drawPolygon[bottom][polygonEFDC]{DGE, CDG, EFD} -are parallelograms on the same base and between the same parallels, and therefore equal. \bycref{prop:I.XXXV} +are parallelograms on the same base and between the same parallels, and therefore equal. \byref{prop:I.XXXV} $\therefore \left\{ \begin{aligned} \polygonABDC &= \mbox{ twice } \polygonBCD\\ \polygonEFDC &= \mbox{ twice } \polygonCDE\\ \end{aligned} - \right\}$ \bycref{prop:I.XXXIV} + \right\}$ \byref{prop:I.XXXIV} $\therefore \polygonBCD = \polygonCDE$. \end{center} @@ -2960,13 +2960,13 @@ \chapter*{Propositions} \mbox{Draw } \drawUnitLine{AC} &\parallel \drawUnitLine{BD}\\ \mbox{and } \drawUnitLine{FH} &\parallel \drawUnitLine{EG}\\ \end{aligned} - \right\}\mbox{\bycref{prop:I.XXXI}}$\\ -$\drawPolygon[bottom][polygonABDC]{ABC, BCD} = \drawPolygon[bottom][polygonEFHG]{EFH, EGH}$ \bycref{prop:I.XXXVI}; + \right\}\mbox{\byref{prop:I.XXXI}}$\\ +$\drawPolygon[bottom][polygonABDC]{ABC, BCD} = \drawPolygon[bottom][polygonEFHG]{EFH, EGH}$ \byref{prop:I.XXXVI}; -but $\polygonABDC = \mbox{ twice } \polygonBCD$ \bycref{prop:I.XXXIV},\\ -and $\polygonEFHG = \mbox{ twice } \polygonEGH$ \bycref{prop:I.XXXIV}, +but $\polygonABDC = \mbox{ twice } \polygonBCD$ \byref{prop:I.XXXIV},\\ +and $\polygonEFHG = \mbox{ twice } \polygonEGH$ \byref{prop:I.XXXIV}, -$\therefore \polygonBCD = \polygonEGH$ \bycref{ax:I.VII}. +$\therefore \polygonBCD = \polygonEGH$ \byref{ax:I.VII}. \end{center} \qed @@ -3002,14 +3002,14 @@ \chapter*{Propositions} on the same base (\drawUnitLine{CD}) and on the same side of it, are between the same parallels.} \begin{center} -If \drawUnitLine{AB}, which joins the vertices of the triangles, be not $\parallel \drawUnitLine{CD}$, draw $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXXI}, meeting \drawUnitLine{CG}. +If \drawUnitLine{AB}, which joins the vertices of the triangles, be not $\parallel \drawUnitLine{CD}$, draw $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXXI}, meeting \drawUnitLine{CG}. Draw \drawUnitLine{DF}. -$\because \drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \bycref{\constref}\\ %Because +$\because \drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \byref{\constref}\\ %Because $\polygonADC = -\drawPolygon[bottom][polygonFDC]{BED, ECD, FBD}$ \bycref{prop:I.XXXVII};\\ -but $\polygonADC = \polygonBDC$ \bycref{\hypref}; +\drawPolygon[bottom][polygonFDC]{BED, ECD, FBD}$ \byref{prop:I.XXXVII};\\ +but $\polygonADC = \polygonBDC$ \byref{\hypref}; $\therefore \polygonBDC = \polygonFDC$, a part equal to the whole, which is absurd. @@ -3063,11 +3063,11 @@ \chapter*{Propositions} \begin{center} If \drawSizedLine{AB} which joins the vertices of triangles be not $\parallel \drawSizedLine{CD,DE,EF}$,\\ -draw \drawSizedLine{AG} $\parallel \drawSizedLine{CD,DE,EF}$ \bycref{prop:I.XXXI},\\ +draw \drawSizedLine{AG} $\parallel \drawSizedLine{CD,DE,EF}$ \byref{prop:I.XXXI},\\ meeting \drawSizedLine{EH}.\\ Draw \drawSizedLine{FG}. -$\because \drawSizedLine{AG} \parallel \drawSizedLine{CD,DE,EF}$ \bycref{\constref}\\ %Because +$\because \drawSizedLine{AG} \parallel \drawSizedLine{CD,DE,EF}$ \byref{\constref}\\ %Because $\polygonACD = \drawFromCurrentPicture[bottom][polygonGEF]{ startAutoLabeling; @@ -3115,8 +3115,8 @@ \chapter*{Propositions} Draw \drawUnitLine{AE} the diagonal. Then -$\drawPolygon[bottom][polygonAED]{AGD,DEG} = \polygonCED$ \bycref{prop:I.XXXVII}\\ -$\polygonABED = \mbox{ twice } \polygonAED$ \bycref{prop:I.XXXIV} +$\drawPolygon[bottom][polygonAED]{AGD,DEG} = \polygonCED$ \byref{prop:I.XXXVII}\\ +$\polygonABED = \mbox{ twice } \polygonAED$ \byref{prop:I.XXXIV} $\therefore \polygonABED = \mbox{ twice } \polygonCED$. \end{center} @@ -3163,22 +3163,22 @@ \chapter*{Propositions} \drawPolygon[bottom][polygonCDG]{DEF,CFE,ECG} and having an angle equal to a given rectilinear angle \drawAngle{I}.} \begin{center} -Make $\drawUnitLine{DE} = \drawUnitLine{EG}$ \bycref{prop:I.X}\\ +Make $\drawUnitLine{DE} = \drawUnitLine{EG}$ \byref{prop:I.X}\\ Draw \drawUnitLine{CE}. -Make $\drawAngle{E} = \drawAngle{I}$ \bycref{prop:I.XXIII}\\ +Make $\drawAngle{E} = \drawAngle{I}$ \byref{prop:I.XXIII}\\ Draw $\left\{ \begin{aligned} \drawUnitLine{AD} &\parallel \drawUnitLine{BE}\\ \drawUnitLine{AC} &\parallel \drawUnitLine{DE}\\ \end{aligned} - \right\}$ \bycref{prop:I.XXXI} + \right\}$ \byref{prop:I.XXXI} $\drawPolygon[bottom][polygonABED]{ABFD,DEF} = \mbox{ twice } -\drawPolygon[bottom][polygonCED]{DEF,CFE}$ \bycref{prop:I.XLI}\\ +\drawPolygon[bottom][polygonCED]{DEF,CFE}$ \byref{prop:I.XLI}\\ but $\polygonCED = -\drawPolygon[bottom][polygonDCG]{ECG}$ \bycref{prop:I.XXXVIII} +\drawPolygon[bottom][polygonDCG]{ECG}$ \byref{prop:I.XXXVIII} $\therefore \polygonABED = \polygonCDG$. \end{center} @@ -3219,10 +3219,10 @@ \chapter*{Propositions} of the parallelograms which are about the diagonal of a parallelogram are equal.} \begin{center} -$\drawPolygon[bottom][polygonADC]{AEF,FCIE,IDE} = \drawPolygon[bottom][polygonABD]{AEH,HBGE,GDE}$ \bycref{prop:I.XXXIV}\\ -and $\drawPolygon[bottom][polygonAEFpIDE]{AEF,IDE} = \drawPolygon[bottom][polygonAEHpGDE]{AEH,GDE}$ \bycref{prop:I.XXXIV} +$\drawPolygon[bottom][polygonADC]{AEF,FCIE,IDE} = \drawPolygon[bottom][polygonABD]{AEH,HBGE,GDE}$ \byref{prop:I.XXXIV}\\ +and $\drawPolygon[bottom][polygonAEFpIDE]{AEF,IDE} = \drawPolygon[bottom][polygonAEHpGDE]{AEH,GDE}$ \byref{prop:I.XXXIV} -$\therefore \polygonFCIE = \polygonHBGE$ \bycref{ax:I.III}. +$\therefore \polygonFCIE = \polygonHBGE$ \byref{ax:I.III}. \end{center} \qed @@ -3284,19 +3284,19 @@ \chapter*{Propositions} \problem[4]{T}{o}{a given straight line (\drawUnitLine{EG}) to apply a parallelogram equal to a given triangle (\drawPolygon[middle][polygonJKL]{JKL}), and having an angle equal to a given rectilinear angle (\drawAngle{N}).} \begin{center} -Make $\drawPolygon[middle][polygonAHEF]{AHEF} = \polygonJKL$ with $\drawAngle{F} = \drawAngle{N}$ \bycref{prop:I.XLII}\\ +Make $\drawPolygon[middle][polygonAHEF]{AHEF} = \polygonJKL$ with $\drawAngle{F} = \drawAngle{N}$ \byref{prop:I.XLII}\\ and having one of its sides \drawUnitLine{FE} conterminous with and in continuation of \drawUnitLine{EG}. Produce \drawUnitLine{AH} till it meets $\drawUnitLine{BG} \parallel \drawUnitLine{HE}$\\ draw \drawUnitLine{BE} produce it till it meets \drawUnitLine{AF} continued;\\ draw $\drawUnitLine{CD} \parallel \drawUnitLine{FE,EG}$ meeting \drawUnitLine{BG} produced and produce \drawUnitLine{HE}. -$\polygonAHEF = \drawPolygon[middle][polygonEGDI]{EGDI}$ \bycref{prop:I.XLIII}\\ -but $\polygonAHEF = \polygonJKL$ \bycref{\constref} +$\polygonAHEF = \drawPolygon[middle][polygonEGDI]{EGDI}$ \byref{prop:I.XLIII}\\ +but $\polygonAHEF = \polygonJKL$ \byref{\constref} $\therefore \polygonEGDI = \polygonJKL$; -and $\drawAngle{F} = \drawAngle{E} =\drawAngle{I} = \drawAngle{N}$ \bycref{prop:I.XXIX,\constref}. % improvement: proposition 19 in the original seems to be a typo +and $\drawAngle{F} = \drawAngle{E} =\drawAngle{I} = \drawAngle{N}$ \byref{prop:I.XXIX,\constref}. % improvement: proposition 19 in the original seems to be a typo \end{center} \qed @@ -3367,17 +3367,17 @@ \chapter*{Propositions} Draw \drawUnitLine{AD} and \drawUnitLine{AC} dividing the rectilinear figure into triangles. Construct $\drawPolygon{FGIH} = \drawPolygon{ADE}$\\ -having $\drawAngle{I} = \drawAngle{O}$ \bycref{prop:I.XLII} +having $\drawAngle{I} = \drawAngle{O}$ \byref{prop:I.XLII} to \drawUnitLine{GI} apply $\drawPolygon{GJKI} = \drawPolygon{ACD}$\\ -having $\drawAngle{K} = \drawAngle{O}$ \bycref{prop:I.XLIV} +having $\drawAngle{K} = \drawAngle{O}$ \byref{prop:I.XLIV} to \drawUnitLine{JK} apply $\drawPolygon{JLMK} = \drawPolygon{ABC}$\\ -having $\drawAngle{M} = \drawAngle{O}$ \bycref{prop:I.XLIV} +having $\drawAngle{M} = \drawAngle{O}$ \byref{prop:I.XLIV} $\therefore \drawPolygon[middle][polygonFLMH]{FGIH,GJKI,JLMK} = \polygonABCDE$ -and \polygonFLMH is a parallelogram. \bycref{prop:I.XXIX,prop:I.XIV,prop:I.XXX}\\ +and \polygonFLMH is a parallelogram. \byref{prop:I.XXIX,prop:I.XIV,prop:I.XXX}\\ having $\drawAngle{M} = \drawAngle{O}$. \end{center} @@ -3408,7 +3408,7 @@ \chapter*{Propositions} \problem{U}{pon}{a given straight line (\drawUnitLine{DC}) to construct a square.} \begin{center} -Draw $\drawUnitLine{CA} \perp \mbox{ and } = \drawUnitLine{DC}$ \bycref{prop:I.XI,prop:I.III} +Draw $\drawUnitLine{CA} \perp \mbox{ and } = \drawUnitLine{DC}$ \byref{prop:I.XI,prop:I.III} Draw $\drawUnitLine{AB} \parallel \drawUnitLine{DC}$,\\ and meeting \drawUnitLine{BD} drawn $\parallel \drawUnitLine{CA}$. @@ -3421,13 +3421,13 @@ \chapter*{Propositions} draw byLabelsOnPolygon(A, B, D, C)(ALL_LABELS, 0); stopTempAngleScale; } -$\drawUnitLine{CA} = \drawUnitLine{DC}$ \bycref{\constref}\\ -$\drawAngle{C} = \mbox{a right angle}$ \bycref{\constref} +$\drawUnitLine{CA} = \drawUnitLine{DC}$ \byref{\constref}\\ +$\drawAngle{C} = \mbox{a right angle}$ \byref{\constref} -$\therefore \drawAngle{D} = \drawAngle{C} = \mbox{a right angle}$ \bycref{prop:I.XXIX},\\ -and~the~remaining sides and~angles must be equal \bycref{prop:I.XXXIV}. +$\therefore \drawAngle{D} = \drawAngle{C} = \mbox{a right angle}$ \byref{prop:I.XXIX},\\ +and~the~remaining sides and~angles must be equal \byref{prop:I.XXXIV}. -And $\therefore \polygonABDC$ is a square \bycref{def:I.XXX}. % improvement: in the original definition 27 was wrongly referenced +And $\therefore \polygonABDC$ is a square \byref{def:I.XXX}. % improvement: in the original definition 27 was wrongly referenced \end{center} \qed @@ -3503,9 +3503,9 @@ \chapter*{Propositions} \problem[4]{I}{n}{a right angled triangle \drawLine[bottom][triangleABC]{CA,BC,AB} the square on the hypotenuse \drawUnitLine{BC} is equal to the sum of the squares of the sides (\drawUnitLine{CA} and \drawUnitLine{AB}).} \begin{center} -On \drawUnitLine{BC}, \drawUnitLine{CA}, \drawUnitLine{AB} describe squares, \bycref{prop:I.XLVI} +On \drawUnitLine{BC}, \drawUnitLine{CA}, \drawUnitLine{AB} describe squares, \byref{prop:I.XLVI} -Draw $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \bycref{prop:I.XXXI}\\ +Draw $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \byref{prop:I.XXXI}\\ also draw \drawUnitLine{BF} and \drawUnitLine{AI}.\\ $\drawAngle{BCI} = \drawAngle{FCA}$. @@ -3566,22 +3566,22 @@ \chapter*{Propositions} \problem{I}{f}{the square of one side (\drawUnitLine{BC}) of a triangle is equal to the squares of the other two sides (\drawUnitLine{AB} and \drawUnitLine{AC}), the angle (\drawAngle{BAC}) subtended by that side is a right angle.} % improvement: DAB sould be BAC https://github.com/jemmybutton/byrne-euclid/issues/31#issuecomment-350511743 \begin{center} -Draw $\drawUnitLine{AD} \perp \drawUnitLine{AB}$ and $= \drawUnitLine{AC}$ \bycref{prop:I.XI,prop:I.III}\\ +Draw $\drawUnitLine{AD} \perp \drawUnitLine{AB}$ and $= \drawUnitLine{AC}$ \byref{prop:I.XI,prop:I.III}\\ and draw \drawUnitLine{BD} also. -Since $\drawUnitLine{AD} = \drawUnitLine{AC}$ \bycref{\constref}\\ +Since $\drawUnitLine{AD} = \drawUnitLine{AC}$ \byref{\constref}\\ $\drawUnitLine{AD}^2 = \drawUnitLine{AC}^2$; $\therefore \drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{AC}^2 + \drawUnitLine{AB}^2$ -but $\drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BD}^2$ \bycref{prop:I.XLVII},\\ -and $\drawUnitLine{AC}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BC}^2$ \bycref{\hypref} +but $\drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BD}^2$ \byref{prop:I.XLVII},\\ +and $\drawUnitLine{AC}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BC}^2$ \byref{\hypref} $\therefore \drawUnitLine{BD}^2 = \drawUnitLine{BC}^2$, $\therefore \drawUnitLine{BD} = \drawUnitLine{BC}$; -and $\therefore \drawAngle{DAB} = \drawAngle{BAC}$ \bycref{prop:I.VIII}, +and $\therefore \drawAngle{DAB} = \drawAngle{BAC}$ \byref{prop:I.VIII}, consequently \drawAngle{BAC} is a right angle. \end{center} @@ -3697,7 +3697,7 @@ \part{Book II} is equal to the sum of the rectangles contained by the undivided line, and the several parts of the divided line.} \begin{center} -Draw $\drawProportionalLine{GB} \perp \drawProportionalLine{GK,KL,LH} \mbox{ and} = \drawProportionalLine{GB}$ \bycref{prop:I.XI,prop:I.III}; complete the parallelograms, that is to say, % improvement: in the original \bycref{prop:I.II} was referenced instead of \bycref{prop:I.XI} +Draw $\drawProportionalLine{GB} \perp \drawProportionalLine{GK,KL,LH} \mbox{ and} = \drawProportionalLine{GB}$ \byref{prop:I.XI,prop:I.III}; complete the parallelograms, that is to say, % improvement: in the original \byref{prop:I.II} was referenced instead of \byref{prop:I.XI} draw $\left\{ \begin{aligned} @@ -3707,7 +3707,7 @@ \part{Book II} \nointerlineskip\hbox{\drawProportionalLine{LE}} \nointerlineskip\hbox{\drawProportionalLine{HC}}} &\parallel \drawProportionalLine{GB}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI} +\right\}$ \byref{prop:I.XXXI} $\drawPolygon[bottom]{GKDB,KLED,LHCE} = \drawPolygon[bottom]{GKDB} + @@ -3758,9 +3758,9 @@ \part{Book II} } \begin{center} -Describe \drawPolygon[bottom][polygonABED]{ACFD,CBEF} \bycref{prop:I.XLVI} +Describe \drawPolygon[bottom][polygonABED]{ACFD,CBEF} \byref{prop:I.XLVI} -Draw \drawProportionalLine{CF} parallel to \drawProportionalLine{AD} \bycref{prop:I.XXXI} +Draw \drawProportionalLine{CF} parallel to \drawProportionalLine{AD} \byref{prop:I.XXXI} $\polygonABED = \drawProportionalLine{AC,CB}^2$ @@ -3803,9 +3803,9 @@ \part{Book II} $\drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DF} = \drawProportionalLine{DF}^2 + \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$.} \begin{center} -Describe \drawPolygon[bottom]{CBED} \bycref{prop:I.XLVI} +Describe \drawPolygon[bottom]{CBED} \byref{prop:I.XLVI} -Describe \drawPolygon[bottom]{ACDF} \bycref{prop:I.XXXI} +Describe \drawPolygon[bottom]{ACDF} \byref{prop:I.XXXI} Then $\drawPolygon[bottom][polygonABEF]{ACDF,CBED} = \polygonCBED + \polygonACDF$, but\\ $\polygonABEF = \drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DE}$ and\\ @@ -3865,21 +3865,21 @@ \part{Book II} } \begin{center} -Describe \drawLine[bottom][squareABED]{AD,BA,KB,EK,FE,DF} \bycref{prop:I.XLVI} +Describe \drawLine[bottom][squareABED]{AD,BA,KB,EK,FE,DF} \byref{prop:I.XLVI} -draw \drawProportionalLine{BG,GD} \bycref{post:I.I}\\ +draw \drawProportionalLine{BG,GD} \byref{post:I.I}\\ and $\left\{ \begin{aligned} \drawProportionalLine{FG,CG} &\parallel \drawProportionalLine{EK,KB}\\ \drawProportionalLine{HG,GK} &\parallel \drawProportionalLine{DF,FE}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}\\ -$\drawAngle{GBK} = \drawAngle{FDG}$ \bycref{prop:I.V},\\ -$\drawAngle{GBK} = \drawAngle{FGD}$ \bycref{prop:I.XXIX}, +\right\}$ \byref{prop:I.XXXI}\\ +$\drawAngle{GBK} = \drawAngle{FDG}$ \byref{prop:I.V},\\ +$\drawAngle{GBK} = \drawAngle{FGD}$ \byref{prop:I.XXIX}, $\therefore \drawAngle{FDG} = \drawAngle{FGD}$ -$\therefore$ by \bycref{prop:I.VI,prop:I.XXIX,prop:I.XXXIV}\\ $\drawFromCurrentPicture[bottom][squareFGHD]{ +$\therefore$ by \byref{prop:I.VI,prop:I.XXIX,prop:I.XXXIV}\\ $\drawFromCurrentPicture[bottom][squareFGHD]{ draw byNamedPolygon(DHG); draw byNamedLineFull(G, G, 1, 0, 0, -1)(DF); draw byNamedLineFull(D, D, 0, 1, 0, -1)(FG); @@ -3893,7 +3893,7 @@ \part{Book II} draw byLabelsOnPolygon(G, C, B, K)(ALL_LABELS, 0); } is a square $= \drawProportionalLine{GK}^2$, -$\drawPolygon[bottom]{ACGH} = \drawPolygon[middle]{GKEF} = \drawProportionalLine{DF} \cdot \drawProportionalLine{GK}$ \bycref{prop:I.XLIII} +$\drawPolygon[bottom]{ACGH} = \drawPolygon[middle]{GKEF} = \drawProportionalLine{DF} \cdot \drawProportionalLine{GK}$ \byref{prop:I.XLIII} But $\drawFromCurrentPicture[bottom][squareABEDf]{ draw byNamedPolygon(GBC,DHG,ACGH,GKEF); @@ -3951,7 +3951,7 @@ \part{Book II} } \begin{center} -Describe \drawPolygon[middle][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \bycref{prop:I.XLVI},\\ +Describe \drawPolygon[middle][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \byref{prop:I.XLVI},\\ draw \drawProportionalLine{EB}\\ and $\left\{ \begin{aligned} @@ -3959,17 +3959,17 @@ \part{Book II} \drawProportionalLine{LM,KL} & \parallel \drawProportionalLine{BD,DC,CA} \\ \drawProportionalLine{AK} & \parallel \drawProportionalLine{CL,LE}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI} +\right\}$ \byref{prop:I.XXXI} -$\drawPolygon{CLKA} = \drawPolygon{DCLH,BDHM}$ \bycref{prop:I.XXXVI}\\ -$\drawPolygon{MHGF} = \drawPolygon{DCLH}$ \bycref{prop:I.XLIII} +$\drawPolygon{CLKA} = \drawPolygon{DCLH,BDHM}$ \byref{prop:I.XXXVI}\\ +$\drawPolygon{MHGF} = \drawPolygon{DCLH}$ \byref{prop:I.XLIII} -$\therefore \mbox{\bycref{ax:I.II} } \drawPolygon{DCLH,BDHM,MHGF} = \drawPolygon{DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA}$ -but $\drawPolygon{HLEG} = \drawProportionalLine{DC}^2$ \bycref{prop:II.IV} +$\therefore \mbox{\byref{ax:I.II} } \drawPolygon{DCLH,BDHM,MHGF} = \drawPolygon{DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA}$ +but $\drawPolygon{HLEG} = \drawProportionalLine{DC}^2$ \byref{prop:II.IV} -and $\squareCBFE = \drawProportionalLine{BD,DC}^2$ \bycref{\constref} +and $\squareCBFE = \drawProportionalLine{BD,DC}^2$ \byref{\constref} -$\therefore \mbox{\bycref{ax:I.II} } \squareCBFE = \drawPolygon{DCLH,CLKA,HLEG}$ +$\therefore \mbox{\byref{ax:I.II} } \squareCBFE = \drawPolygon{DCLH,CLKA,HLEG}$ $\therefore \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA} + \drawProportionalLine{DC}^2 = \drawProportionalLine{CA}^2 = \drawProportionalLine{BD,DC}^2$. \end{center} @@ -4019,21 +4019,21 @@ \part{Book II} } \begin{center} -Describe \drawPolygon[bottom][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \bycref{prop:I.XLVI}, draw \drawProportionalLine{EB}\\ +Describe \drawPolygon[bottom][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \byref{prop:I.XLVI}, draw \drawProportionalLine{EB}\\ and $\left\{ \begin{aligned} \drawProportionalLine{HG,DH} & \parallel \drawProportionalLine{CL,LE} \\ \drawProportionalLine{LM,KL} & \parallel \drawProportionalLine{BD,DC,CA} \\ \drawProportionalLine{AK} & \parallel \drawProportionalLine{CL,LE}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI} +\right\}$ \byref{prop:I.XXXI} -$\drawPolygon[bottom]{MHGF} = \drawPolygon[bottom]{DCLH} = \drawPolygon[bottom]{CLKA}$ \bycref{prop:I.XXXVI,prop:I.XLIII} +$\drawPolygon[bottom]{MHGF} = \drawPolygon[bottom]{DCLH} = \drawPolygon[bottom]{CLKA}$ \byref{prop:I.XXXVI,prop:I.XLIII} $\therefore \drawPolygon[bottom]{BDHM,DCLH,MHGF} = \drawPolygon[bottom]{BDHM,DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{BD,DC,CA}$;\\ -but $\drawPolygon[bottom]{HLEG} = \drawProportionalLine{DC}^2$ \bycref{prop:II.IV} +but $\drawPolygon[bottom]{HLEG} = \drawProportionalLine{DC}^2$ \byref{prop:II.IV} -$\therefore \squareCBFE = \drawProportionalLine{CB}^2 = \drawPolygon[bottom]{BDHM,DCLH,CLKA,HLEG}$ \bycref{\constref,ax:I.II} +$\therefore \squareCBFE = \drawProportionalLine{CB}^2 = \drawPolygon[bottom]{BDHM,DCLH,CLKA,HLEG}$ \byref{\constref,ax:I.II} $\therefore \drawProportionalLine{BD,DC,CA} \cdot \drawProportionalLine{BD} + \drawProportionalLine{DC}^2 = \drawProportionalLine{BD,DC}^2$. \end{center} @@ -4076,18 +4076,18 @@ \part{Book II} } \begin{center} -Describe \drawPolygon[bottom][squareABED]{DNGH,NEFG,HGCA,GFBC} \bycref{prop:I.XLVI}, draw \drawProportionalLine{BD} \bycref{post:I.I},\\ +Describe \drawPolygon[bottom][squareABED]{DNGH,NEFG,HGCA,GFBC} \byref{prop:I.XLVI}, draw \drawProportionalLine{BD} \byref{post:I.I},\\ and $\left\{ \begin{aligned} \drawProportionalLine{GN,GC} &\parallel \drawProportionalLine{EB} \\ \drawProportionalLine{GH,GF} &\parallel \drawProportionalLine{DN,NE}\\ \end{aligned} \right\}$\\ -$\drawPolygon[bottom]{HGCA} = \drawPolygon[bottom]{NEFG}$ \bycref{prop:I.XLIII},\\ -add $\drawPolygon[bottom]{GFBC} = \drawProportionalLine{NE}^2$ to both \bycref{prop:II.IV} +$\drawPolygon[bottom]{HGCA} = \drawPolygon[bottom]{NEFG}$ \byref{prop:I.XLIII},\\ +add $\drawPolygon[bottom]{GFBC} = \drawProportionalLine{NE}^2$ to both \byref{prop:II.IV} $\drawPolygon[bottom]{HGCA,GFBC} = \drawPolygon[bottom]{NEFG,GFBC} = \drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE}$\\ -$\drawPolygon[bottom]{DNGH} = \drawProportionalLine{DN}^2$ \bycref{prop:II.IV}\\ +$\drawPolygon[bottom]{DNGH} = \drawProportionalLine{DN}^2$ \byref{prop:II.IV}\\ $\drawPolygon[bottom]{HGCA,GFBC} + \drawPolygon[bottom]{NEFG,GFBC} + \drawPolygon[bottom]{DNGH} = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2 = \squareABED + \drawPolygon[bottom]{GFBC}$; $\drawProportionalLine{DN,NE}^2 + \drawProportionalLine{NE}^2 = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2$. @@ -4153,7 +4153,7 @@ \part{Book II} Construct \drawFromCurrentPicture[bottom]{ draw byNamedLine(BL,CH,MN,OP); draw byNamedLineSeq(0)(LF,HL,EH,EA,AD,DF); -} \bycref{prop:I.XLVI};\\ +} \byref{prop:I.XLVI};\\ draw \drawProportionalLine{DE}, $\left. @@ -4174,10 +4174,10 @@ \part{Book II} \end{aligned} \right\} & \parallel \drawProportionalLine{EH,HL,LF}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI} +\right\}$ \byref{prop:I.XXXI} -$\drawProportionalLine{EH,HL,LF}^2 = \drawProportionalLine{LF}^2 + \drawProportionalLine{EH,HL}^2 + 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{LF}$ \bycref{prop:II.IV}\\ -but $\drawProportionalLine{HL}^2 + \drawProportionalLine{EH,HL}^2 = 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$ \bycref{prop:II.VII} +$\drawProportionalLine{EH,HL,LF}^2 = \drawProportionalLine{LF}^2 + \drawProportionalLine{EH,HL}^2 + 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{LF}$ \byref{prop:II.IV}\\ +but $\drawProportionalLine{HL}^2 + \drawProportionalLine{EH,HL}^2 = 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$ \byref{prop:II.VII} $\therefore \drawProportionalLine{EH,HL,LF}^2 = 4 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$. \end{center} @@ -4231,12 +4231,12 @@ \part{Book II} Draw \drawProportionalLine{AE} and \drawProportionalLine{EH,HF,FB},\\ $\drawProportionalLine{FD} \parallel \drawProportionalLine{CG,GE}$, $ \drawProportionalLine{GF} \parallel \drawProportionalLine{CD,DB}$ and draw \drawProportionalLine{AF}. -$\drawAngle{A} = \drawAngle{AEC}$ \bycref{prop:I.V} $=$ half a right angle \bycref{prop:I.XXXII}\\ -$\drawAngle{B} = \drawAngle{DFB}$ \bycref{prop:I.V} $=$ half a right angle \bycref{prop:I.XXXII}\\ +$\drawAngle{A} = \drawAngle{AEC}$ \byref{prop:I.V} $=$ half a right angle \byref{prop:I.XXXII}\\ +$\drawAngle{B} = \drawAngle{DFB}$ \byref{prop:I.V} $=$ half a right angle \byref{prop:I.XXXII}\\ $\therefore \drawAngle{AEC,CEB} =$ a right angle. -$\drawAngle{B} = \drawAngle{CEB} = \drawAngle{EFG} = \drawAngle{DFB}$ \bycref{prop:I.V, prop:I.XXIX}.\\ -hence $\drawProportionalLine{FD} = \drawProportionalLine{DB}$, $\drawProportionalLine{GE} = \drawProportionalLine{GF} = \drawProportionalLine{CD}$ \bycref{prop:I.VI,prop:I.XXXIV} +$\drawAngle{B} = \drawAngle{CEB} = \drawAngle{EFG} = \drawAngle{DFB}$ \byref{prop:I.V, prop:I.XXIX}.\\ +hence $\drawProportionalLine{FD} = \drawProportionalLine{DB}$, $\drawProportionalLine{GE} = \drawProportionalLine{GF} = \drawProportionalLine{CD}$ \byref{prop:I.VI,prop:I.XXXIV} $\drawProportionalLine{AF}^2 = \left\{ \begin{aligned} @@ -4246,7 +4246,7 @@ \part{Book II} &= \drawProportionalLine{AE}^2 + \drawProportionalLine{EH,HF}^2 \\ &= 2 \cdot \drawProportionalLine{AC}^2 + 2 \cdot \drawProportionalLine{CD}^2\\ \end{aligned} - \right. \mbox{ \bycref{prop:I.XLVII}} + \right. \mbox{ \byref{prop:I.XLVII}} \end{aligned}\right.$ $\therefore \drawProportionalLine{AC,CD}^2 + \drawProportionalLine{DB}^2 = 2 \cdot \drawProportionalLine{AC}^2 + 2 \cdot \drawProportionalLine{CD}^2$. @@ -4303,18 +4303,18 @@ \part{Book II} \drawProportionalLine{FD,DG} & \parallel \drawProportionalLine{CE} \\ \drawProportionalLine{EF} & \parallel \drawProportionalLine{CB,BD}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}\\ +\right\}$ \byref{prop:I.XXXI}\\ draw \drawProportionalLine{AG} also. -$\drawAngle{A} = \drawAngle{CEA}$ \bycref{prop:I.V} $=$ half a right angle \bycref{prop:I.XXXII}\\ -$\drawAngle{CBE} = \drawAngle{BEC}$ \bycref{prop:I.V} $=$ half a right angle \bycref{prop:I.XXXII}\\ +$\drawAngle{A} = \drawAngle{CEA}$ \byref{prop:I.V} $=$ half a right angle \byref{prop:I.XXXII}\\ +$\drawAngle{CBE} = \drawAngle{BEC}$ \byref{prop:I.V} $=$ half a right angle \byref{prop:I.XXXII}\\ $\therefore \drawAngle{CEA,BEC} = $ a right angle. $\drawAngle{DBG} = \drawAngle{CBE} = \drawAngle{BEC} = \drawAngle{FEB} = \drawAngle{G} =$\\ -half a right angle \bycref{prop:I.V,prop:I.XXXII,prop:I.XXIX,prop:I.XXXIV},\\ % improvement: in the original DBG is rotated similarly to A; most likely we only need prop:I.XXIX and prop:I.XV here -and $\drawProportionalLine{BD} = \drawProportionalLine{DG}$, $\drawProportionalLine{CB,BD} = \drawProportionalLine{EF} = \drawProportionalLine{FD,DG}$, \bycref{prop:I.VI,prop:I.XXXIV}. +half a right angle \byref{prop:I.V,prop:I.XXXII,prop:I.XXIX,prop:I.XXXIV},\\ % improvement: in the original DBG is rotated similarly to A; most likely we only need prop:I.XXIX and prop:I.XV here +and $\drawProportionalLine{BD} = \drawProportionalLine{DG}$, $\drawProportionalLine{CB,BD} = \drawProportionalLine{EF} = \drawProportionalLine{FD,DG}$, \byref{prop:I.VI,prop:I.XXXIV}. -Hence by \bycref{prop:I.XLVII} +Hence by \byref{prop:I.XLVII} $\drawProportionalLine{AG}^2 = \left\{ \begin{aligned} @@ -4366,15 +4366,15 @@ \part{Book II} } \begin{center} -Describe \drawPolygon[bottom]{AHKC,HBDK} \bycref{prop:I.XLVI},\\ -make $\drawProportionalLine{EA} = \drawProportionalLine{CE}$ \bycref{prop:I.X},\\ +Describe \drawPolygon[bottom]{AHKC,HBDK} \byref{prop:I.XLVI},\\ +make $\drawProportionalLine{EA} = \drawProportionalLine{CE}$ \byref{prop:I.X},\\ draw \drawProportionalLine{BE},\\ -take $\drawProportionalLine{EA,AF} = \drawProportionalLine{BE}$ \bycref{prop:I.III},\\ -on \drawProportionalLine{AF} describe \drawPolygon[bottom]{AHGF} \bycref{prop:I.XLVI}. +take $\drawProportionalLine{EA,AF} = \drawProportionalLine{BE}$ \byref{prop:I.III},\\ +on \drawProportionalLine{AF} describe \drawPolygon[bottom]{AHGF} \byref{prop:I.XLVI}. -Produce \drawProportionalLine{KG} \bycref{post:I.II}. +Produce \drawProportionalLine{KG} \byref{post:I.II}. -Then, \bycref{prop:II.VI} $\drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} + \drawProportionalLine{EA}^2 = \drawProportionalLine{EA,AF}^2 = \drawProportionalLine{BE}^2 = \drawProportionalLine{AH,HB}^2 + \drawProportionalLine{EA}^2 \therefore \drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} = \drawProportionalLine{AH,HB}^2$, or,\\ +Then, \byref{prop:II.VI} $\drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} + \drawProportionalLine{EA}^2 = \drawProportionalLine{EA,AF}^2 = \drawProportionalLine{BE}^2 = \drawProportionalLine{AH,HB}^2 + \drawProportionalLine{EA}^2 \therefore \drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} = \drawProportionalLine{AH,HB}^2$, or,\\ $\drawPolygon[bottom]{AHKC,AHGF} = \drawPolygon[bottom]{AHKC,HBDK} \therefore \drawPolygon[bottom]{AHGF} = \drawPolygon[bottom]{HBDK}$ $\therefore \drawProportionalLine{AH,HB} \cdot \drawProportionalLine{HB} = \drawProportionalLine{AH}^2$. @@ -4407,18 +4407,18 @@ \part{Book II} } \begin{center} -By \bycref{prop:II.IV}\\ +By \byref{prop:II.IV}\\ $\drawProportionalLine{CA,AD}^2 = \drawProportionalLine{CA}^2 + \drawProportionalLine{AD}^2 + 2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD}$: add $\drawProportionalLine{DB}^2$ to both\\ -$\drawProportionalLine{CA,AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{BC}^2$ \bycref{prop:I.XLVII}\\ +$\drawProportionalLine{CA,AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{BC}^2$ \byref{prop:I.XLVII}\\ $= 2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD} + \drawProportionalLine{CA}^2 + \left\{ \begin{aligned} &\drawProportionalLine{AD}^2 \\ &\drawProportionalLine{DB}^2\\ \end{aligned} \right\}$ or\\ -$+ \drawProportionalLine{BA}^2$ \bycref{prop:I.XLVII}. % original presentation is not very clear. this is meant to be AD^2 + DB^2 which is, by I.XLVII = BA^2 +$+ \drawProportionalLine{BA}^2$ \byref{prop:I.XLVII}. % original presentation is not very clear. this is meant to be AD^2 + DB^2 which is, by I.XLVII = BA^2 $\therefore \drawProportionalLine{BC}^2 = 2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD} + \drawProportionalLine{CA}^2 + \drawProportionalLine{BA}^2$: hence $ \drawProportionalLine{BC}^2 > \drawProportionalLine{CA}^2 + \drawProportionalLine{BA}^2$ by $2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD}$. \end{center} @@ -4466,19 +4466,19 @@ \part{Book II} } \begin{center} -First, suppose the perpendicular to fall within the triangle, then \bycref{prop:II.VII}\\ +First, suppose the perpendicular to fall within the triangle, then \byref{prop:II.VII}\\ $\drawProportionalLine{BD,DC}^2 + \drawProportionalLine{BD}^2 = 2 \cdot \drawProportionalLine{BD,DC} \cdot \drawProportionalLine{BD} + \drawProportionalLine{DC}^2$,\\ add to each $\drawProportionalLine{AD}^2$ then,\\ $\drawProportionalLine{BD,DC}^2 + \drawProportionalLine{BD}^2 + \drawProportionalLine{AD}^2 = 2 \cdot \drawProportionalLine{BD,DC} \cdot \drawProportionalLine{BD} + \drawProportionalLine{DC}^2 + \drawProportionalLine{AD}^2$\\ -$\therefore$ \bycref{prop:I.XLVII}\\ +$\therefore$ \byref{prop:I.XLVII}\\ $\drawProportionalLine{BD,DC}^2 + \drawProportionalLine{AB}^2 = 2 \cdot \drawProportionalLine{BD,DC} \cdot \drawProportionalLine{BD} + \drawProportionalLine{CA}^2$,\\ and $\therefore \drawProportionalLine{CA}^2 < \drawProportionalLine{BD,DC}^2 + \drawProportionalLine{AB}^2$ by $2 \cdot \drawProportionalLine{BD,DC} \cdot \drawProportionalLine{BD}$ -Next suppose the perpendicular to fall without the triangle, then \bycref{prop:II.VII}\\ +Next suppose the perpendicular to fall without the triangle, then \byref{prop:II.VII}\\ $\drawProportionalLine{FG,GH}^2 + \drawProportionalLine{FG}^2 = 2 \cdot \drawProportionalLine{FG,GH} \cdot \drawProportionalLine{FG} + \drawProportionalLine{GH}^2$,\\ add to each $\drawProportionalLine{HE}^2$ then\\ $\drawProportionalLine{FG,GH}^2 + \drawProportionalLine{FG}^2 + \drawProportionalLine{HE}^2= 2 \cdot \drawProportionalLine{FG,GH} \cdot \drawProportionalLine{FG} + \drawProportionalLine{GH}^2 + \drawProportionalLine{HE}^2$\\ -$\therefore$ \bycref{prop:I.XLVII}\\ +$\therefore$ \byref{prop:I.XLVII}\\ $\drawProportionalLine{EF} + \drawProportionalLine{FG}^2 = 2 \cdot \drawProportionalLine{FG,GH} \cdot \drawProportionalLine{FG}^2 + \drawProportionalLine{EG}^2$,\\ $\therefore \drawProportionalLine{EG}^2 < \drawProportionalLine{EF}^2 + \drawProportionalLine{FG}^2$ by $2 \cdot \drawProportionalLine{FG,GH} \cdot \drawProportionalLine{FG}$. \end{center} @@ -4541,9 +4541,9 @@ \part{Book II} } \begin{center} -Make $\drawPolygon{BCDE} = \drawPolygon{abcdef}$ \bycref{prop:I.XLV}, +Make $\drawPolygon{BCDE} = \drawPolygon{abcdef}$ \byref{prop:I.XLV}, produce \drawSizedLine{EG,GB} until $\drawSizedLine{FE} = \drawSizedLine{ED}$;\\ -take $\drawSizedLine{FE,EG} = \drawSizedLine{GB}$ \bycref{prop:I.X}. +take $\drawSizedLine{FE,EG} = \drawSizedLine{GB}$ \byref{prop:I.X}. Describe \drawFromCurrentPicture[bottom]{ @@ -4554,11 +4554,11 @@ \part{Book II} stopAutoLabeling; stopTempScale; } -\bycref{post:I.III},\\ +\byref{post:I.III},\\ and produce \drawSizedLine{ED} to meet it: draw \drawSizedLine{HG}. -$\drawSizedLine{BG}^2 \mbox{ or } \drawSizedLine{HG}^2 = \drawSizedLine{FE} \cdot \drawSizedLine{EG,GB} + \drawSizedLine{EG}^2$ \bycref{prop:II.V},\\ -but $\drawSizedLine{HG}^2 = \drawSizedLine{HE} ^2 + \drawSizedLine{EG}^2$ \bycref{prop:I.XLVII}; +$\drawSizedLine{BG}^2 \mbox{ or } \drawSizedLine{HG}^2 = \drawSizedLine{FE} \cdot \drawSizedLine{EG,GB} + \drawSizedLine{EG}^2$ \byref{prop:II.V},\\ +but $\drawSizedLine{HG}^2 = \drawSizedLine{HE} ^2 + \drawSizedLine{EG}^2$ \byref{prop:I.XLVII}; $\therefore \drawSizedLine{HE}^2 + \drawSizedLine{EG}^2 = \drawSizedLine{FE} \cdot \drawSizedLine{EG,GB} + \drawSizedLine{EG}^2$ @@ -4790,7 +4790,7 @@ \chapter*{Definitions} For, if it be possible, let any other point as the point of concourse of \drawUnitLine{AG}, \drawUnitLine{DG} and \drawUnitLine{BG} be the centre. -Because in \drawLine[bottom][triangleADG]{AG,DG,AD} and \drawLine[middle][triangleDGB]{DB,DG,BG} $\drawUnitLine{AG} = \drawUnitLine{BG}$ \bycref{\hypref,def:I.XV}, $\drawUnitLine{AD} = \drawUnitLine{DB}$ \bycref{\constref} and \drawUnitLine{DG} common, $\drawAngle{ADF,FDG} = \drawAngle{GDB}$ \bycref{prop:I.VIII}, and are therefore right angles; but $\drawAngle{FDG,GDB} = \drawRightAngle$ \bycref{\constref} $\drawAngle{GDB} = \drawAngle{FDG,GDB}$ \bycref{ax:I.XI} which is absurd; therefore the assumed point is not the centre of the circle; and in the same manner it can be proved that no other point which is not on \drawUnitLine{EC} is the centre, therefore the centre is in \drawUnitLine{EC}, and therefore the point where \drawUnitLine{EC} is bisected is the centre. +Because in \drawLine[bottom][triangleADG]{AG,DG,AD} and \drawLine[middle][triangleDGB]{DB,DG,BG} $\drawUnitLine{AG} = \drawUnitLine{BG}$ \byref{\hypref,def:I.XV}, $\drawUnitLine{AD} = \drawUnitLine{DB}$ \byref{\constref} and \drawUnitLine{DG} common, $\drawAngle{ADF,FDG} = \drawAngle{GDB}$ \byref{prop:I.VIII}, and are therefore right angles; but $\drawAngle{FDG,GDB} = \drawRightAngle$ \byref{\constref} $\drawAngle{GDB} = \drawAngle{FDG,GDB}$ \byref{ax:I.XI} which is absurd; therefore the assumed point is not the centre of the circle; and in the same manner it can be proved that no other point which is not on \drawUnitLine{EC} is the centre, therefore the centre is in \drawUnitLine{EC}, and therefore the point where \drawUnitLine{EC} is bisected is the centre. \qed @@ -4822,14 +4822,14 @@ \chapter*{Definitions} \problem[2]{A}{ straight}{line (\drawSizedLine{AB}) joining two points in the circumference of a circle \drawCircle[middle][1/4]{D}, lies wholly within the circle.} \begin{center} -Find the centre of \circleD\ \bycref{prop:III.I};\\ +Find the centre of \circleD\ \byref{prop:III.I};\\ from the centre draw \drawSizedLine{DE} to any point in \drawSizedLine{AB}, meeting the circumference from the centre;\\ draw \drawSizedLine{AD} and \drawSizedLine{BD}. -Then $\drawAngle{A} = \drawAngle{B}$ \bycref{prop:I.V}\\ -but $\drawAngle{E} > \drawAngle{A} \mbox{ or } > \drawAngle{B}$ \bycref{prop:I.XVI} +Then $\drawAngle{A} = \drawAngle{B}$ \byref{prop:I.V}\\ +but $\drawAngle{E} > \drawAngle{A} \mbox{ or } > \drawAngle{B}$ \byref{prop:I.XVI} -$\therefore \drawSizedLine{AD} > \drawSizedLine{DE}$ \bycref{prop:I.XIX}\\ +$\therefore \drawSizedLine{AD} > \drawSizedLine{DE}$ \byref{prop:I.XIX}\\ but $\drawSizedLine{AD} = \drawSizedLine{DE,EF}$, $\therefore \drawSizedLine{DE,EF} > \drawSizedLine{DE}$; @@ -4880,16 +4880,16 @@ \chapter*{Definitions} In \drawLine[bottom][triangleAEF]{AE,EF,AF} and \drawLine[bottom][triangleFEB]{EF,EB,FB}\\ $\drawUnitLine{AE} = \drawUnitLine{EB}$, \drawUnitLine{EF} common,\\ -and $\drawUnitLine{AF} = \drawUnitLine{FB} \therefore \drawAngle{AFE} = \drawAngle{EFB}$ \bycref{prop:I.VIII}\\ -and $\therefore \drawUnitLine{EF} \perp \drawUnitLine{AF,FB}$ \bycref{def:I.X} % improvement: definition 7 in the original is irrelevant +and $\drawUnitLine{AF} = \drawUnitLine{FB} \therefore \drawAngle{AFE} = \drawAngle{EFB}$ \byref{prop:I.VIII}\\ +and $\therefore \drawUnitLine{EF} \perp \drawUnitLine{AF,FB}$ \byref{def:I.X} % improvement: definition 7 in the original is irrelevant Again let $\drawUnitLine{EF} \perp \drawUnitLine{AF,FB}$\\ Then in \triangleAEF\ and \triangleFEB\\ -$\drawAngle{A} = \drawAngle{B}$ \bycref{prop:I.V}\\ -$\drawAngle{AFE} = \drawAngle{EFB}$ \bycref{\hypref}\\ +$\drawAngle{A} = \drawAngle{B}$ \byref{prop:I.V}\\ +$\drawAngle{AFE} = \drawAngle{EFB}$ \byref{\hypref}\\ and $\drawUnitLine{AE} = \drawUnitLine{EB}$ -$\therefore \drawUnitLine{AF} = \drawUnitLine{FB}$ \bycref{prop:I.XXVI} +$\therefore \drawUnitLine{AF} = \drawUnitLine{FB}$ \byref{prop:I.XXVI} and $\therefore \drawUnitLine{EF}$ bisects \drawUnitLine{AF,FB}. \end{center} @@ -4926,9 +4926,9 @@ \chapter*{Definitions} But if neither of the lines \drawUnitLine{AC} or \drawUnitLine{BD} pass through the centre, draw \drawUnitLine{EF} from the centre to their intersection. \begin{center} -If \drawUnitLine{AC} be bisected, \drawUnitLine{EF} $\perp$ to it \bycref{prop:III.III}\\ +If \drawUnitLine{AC} be bisected, \drawUnitLine{EF} $\perp$ to it \byref{prop:III.III}\\ $\therefore \drawAngle{FED,DEC} = \drawRightAngle$\\ -and if \drawUnitLine{BD} be bisected, $\drawUnitLine{EF} \perp \drawUnitLine{BD}$ \bycref{prop:III.III}\\ +and if \drawUnitLine{BD} be bisected, $\drawUnitLine{EF} \perp \drawUnitLine{BD}$ \byref{prop:III.III}\\ $\therefore \drawAngle{FED} = \drawRightAngle$; and $\therefore \drawAngle{FED} = \drawAngle{FED,DEC}$;\\ @@ -4974,8 +4974,8 @@ \chapter*{Definitions} Suppose it possible that two intersecting circles have a common centre; from such supposed centre draw \drawUnitLine{CE} to the intersecting point, and \drawUnitLine{EF,FG} meeting the circumferences of the circles. \begin{center} -Then $\drawUnitLine{CE} = \drawUnitLine{EF}$ \bycref{def:I.XV}\\ -and $\drawUnitLine{CE} = \drawUnitLine{EF,FG}$ \bycref{def:I.XV} +Then $\drawUnitLine{CE} = \drawUnitLine{EF}$ \byref{def:I.XV}\\ +and $\drawUnitLine{CE} = \drawUnitLine{EF,FG}$ \byref{def:I.XV} $\therefore \drawUnitLine{EF} = \drawUnitLine{EF,FG}$\\ a part equal to the whole, which is absurd: @@ -5018,8 +5018,8 @@ \chapter*{Definitions} For, if it be possible, let both circles have the same centre; from such a supposed centre draw \drawUnitLine{FE,EB}, and \drawUnitLine{CF} to the point of contact. \begin{center} -Then $\drawUnitLine{CF} = \drawUnitLine{FE}$ \bycref{def:I.XV}\\ -and $\drawUnitLine{CF} = \drawUnitLine{FE,EB}$ \bycref{def:I.XV} +Then $\drawUnitLine{CF} = \drawUnitLine{FE}$ \byref{def:I.XV}\\ +and $\drawUnitLine{CF} = \drawUnitLine{FE,EB}$ \byref{def:I.XV} $\therefore \drawUnitLine{FE} = \drawUnitLine{FE,EB}$; \end{center} @@ -5097,7 +5097,7 @@ \chapter*{Definitions} Fig. 2. The two lines (\drawUnitLine{FM,MK} and \drawUnitLine{FG}) which make equal angles with that passing through the centre, on opposite sides of it, are equal to each other; and there cannot be drawn a third line equal to them, from the same point to the circumference.} \startsubproposition{Figure I.} -To the centre of the circle draw \drawUnitLine{EB} and \drawUnitLine{EC}; then $\drawUnitLine{AE} = \drawUnitLine{EB}$ \bycref{def:I.XV} $\drawUnitLine{EF,AE} = \drawUnitLine{EF} + \drawUnitLine{EB} > \drawUnitLine{FB}$ \bycref{prop:I.XX} in like manner \drawUnitLine{EF,AE} may be shown to be greater than \drawUnitLine{FC}, or any other line drawn from the same point to the circumference. Again, by \bycref{prop:I.XX} $\drawUnitLine{EF} + \drawUnitLine{FC} > \drawUnitLine{EC} = \drawUnitLine{FD} + \drawUnitLine{EF}$, take \drawUnitLine{EF} from both; $\therefore \drawUnitLine{FC} > \drawUnitLine{FD}$ \bycref{ax:I.III}, and in like manner it may be shown that \drawUnitLine{FD} is less than any other line drawn from the same point to the circumference. Again, in \drawLine[middle][triangleEFB]{FB,EF,EB} and \drawLine[middle][triangleEFC]{FC,EF,EC}, \drawUnitLine{EF} common, $\drawAngle{BEC,CEF} > \drawAngle{CEF}$, and $\drawUnitLine{EB} = \drawUnitLine{EC} \therefore \drawUnitLine{FB} > \drawUnitLine{FC}$ \bycref{prop:I.XXIV} and \drawUnitLine{FB} may in like manner to be proved greater than any other line drawn from the same point to the circumference more remote from \drawUnitLine{EF,AE}. +To the centre of the circle draw \drawUnitLine{EB} and \drawUnitLine{EC}; then $\drawUnitLine{AE} = \drawUnitLine{EB}$ \byref{def:I.XV} $\drawUnitLine{EF,AE} = \drawUnitLine{EF} + \drawUnitLine{EB} > \drawUnitLine{FB}$ \byref{prop:I.XX} in like manner \drawUnitLine{EF,AE} may be shown to be greater than \drawUnitLine{FC}, or any other line drawn from the same point to the circumference. Again, by \byref{prop:I.XX} $\drawUnitLine{EF} + \drawUnitLine{FC} > \drawUnitLine{EC} = \drawUnitLine{FD} + \drawUnitLine{EF}$, take \drawUnitLine{EF} from both; $\therefore \drawUnitLine{FC} > \drawUnitLine{FD}$ \byref{ax:I.III}, and in like manner it may be shown that \drawUnitLine{FD} is less than any other line drawn from the same point to the circumference. Again, in \drawLine[middle][triangleEFB]{FB,EF,EB} and \drawLine[middle][triangleEFC]{FC,EF,EC}, \drawUnitLine{EF} common, $\drawAngle{BEC,CEF} > \drawAngle{CEF}$, and $\drawUnitLine{EB} = \drawUnitLine{EC} \therefore \drawUnitLine{FB} > \drawUnitLine{FC}$ \byref{prop:I.XXIV} and \drawUnitLine{FB} may in like manner to be proved greater than any other line drawn from the same point to the circumference more remote from \drawUnitLine{EF,AE}. \startsubproposition{Figure II.} \begin{center} @@ -5106,7 +5106,7 @@ \chapter*{Definitions} then in \drawLine[middle][triangleEFM]{EM,EF,FM} and \drawLine[middle][triangleEFG]{FG,EF,EG}, \drawUnitLine{EF} common,\\ $\drawAngle{KFE} = \drawAngle{GFE}$ and $\drawUnitLine{FG} = \drawUnitLine{FM}$ -$\therefore \drawUnitLine{EG} = \drawUnitLine{EM}$ \bycref{prop:I.IV} +$\therefore \drawUnitLine{EG} = \drawUnitLine{EM}$ \byref{prop:I.IV} $\therefore \drawUnitLine{EG} = \drawUnitLine{EM,MH} = \drawUnitLine{EM}$\\ a part equal to the whole, which is absurd: @@ -5155,10 +5155,10 @@ \chapter*{Definitions} Draw \drawUnitLine{MF} and \drawUnitLine{ME} to the centre.\\ Then, \drawUnitLine{DM,MA} which passes through the centre, is greatest;\\ for since $\drawUnitLine{MA} = \drawUnitLine{ME}$, if \drawUnitLine{DM} be added to both $\drawUnitLine{DM,MA} = \drawUnitLine{DM} + \drawUnitLine{ME}$;\\ -but $> \drawUnitLine{DE}$ \bycref{prop:I.XX}\\ +but $> \drawUnitLine{DE}$ \byref{prop:I.XX}\\ $\therefore$ \drawUnitLine{DM,MA} is greater than any other line drawn from the same point to the concave circumference.\\ Again in \drawLine[middle][triangleDFM]{DM,MF,DF} and \drawLine[middle][triangleDEM]{DM,ME,DE}, $\drawUnitLine{MF} = \drawUnitLine{ME}$, and \drawUnitLine{DM} common,\\ -but $\drawAngle{DMF,FME} > \drawAngle{DMF}$, $\therefore \drawUnitLine{DE} > \drawUnitLine{DF}$ \bycref{prop:I.XXIV};\\ +but $\drawAngle{DMF,FME} > \drawAngle{DMF}$, $\therefore \drawUnitLine{DE} > \drawUnitLine{DF}$ \byref{prop:I.XXIV};\\ and in like manner \drawUnitLine{DE} may be shown $>$ than any other line more remote from \drawUnitLine{DM,MA}. \end{center} @@ -5192,11 +5192,11 @@ \chapter*{Definitions} Of those lines falling on the convex circumference the least is that (\drawUnitLine{DG}) which being produced would pass through the centre, and the line which is nearer to least is less than that which is more remote. \begin{center} -For, since $\drawUnitLine{KM} + \drawUnitLine{DK} > \drawUnitLine{DG,GM}$ \bycref{prop:I.XX}\\ +For, since $\drawUnitLine{KM} + \drawUnitLine{DK} > \drawUnitLine{DG,GM}$ \byref{prop:I.XX}\\ and $\drawUnitLine{KM} = \drawUnitLine{GM}$,\\ -$\therefore \drawUnitLine{DK} > \drawUnitLine{DG}$ \bycref{ax:I.V} +$\therefore \drawUnitLine{DK} > \drawUnitLine{DG}$ \byref{ax:I.V} -And again, since $\drawUnitLine{HM} + \drawUnitLine{DH} > \drawUnitLine{KM} + \drawUnitLine{DK}$ \bycref{prop:I.XXI},\\ +And again, since $\drawUnitLine{HM} + \drawUnitLine{DH} > \drawUnitLine{KM} + \drawUnitLine{DK}$ \byref{prop:I.XXI},\\ and $\drawUnitLine{HM} = \drawUnitLine{KM}$, $\therefore \drawUnitLine{DK} < \drawUnitLine{DH}$. And so of others. @@ -5243,7 +5243,7 @@ \chapter*{Definitions} Then in \drawLine[middle][triangleDMO]{DM,DO,BO,BM} and \drawLine[middle][triangleDMH]{HM,DH,DM} we have $\drawUnitLine{DO} = \drawUnitLine{DH}$\\ and \drawUnitLine{DM} common, and also $\drawAngle{MDN} = \drawAngle{HDM}$, -$\therefore \drawUnitLine{BM,BO} = \drawUnitLine{HM}$ \bycref{prop:I.IV};\\ +$\therefore \drawUnitLine{BM,BO} = \drawUnitLine{HM}$ \byref{prop:I.IV};\\ but $\drawUnitLine{HM} = \drawUnitLine{BM}$;\\ $\therefore \drawUnitLine{BM} = \drawUnitLine{BM,BO}$, which is absurd. @@ -5287,7 +5287,7 @@ \chapter*{Definitions} stopGlobalRotation; } must be; join these two points by \drawUnitLine{DF} and produce it both ways to the circumference. -Then since more than two equal straight lines are drawn from a point which is not the centre, to the circumference, two of them at least must lie at the same side of the diameter \drawUnitLine{DH,DF,FL}; and since from a point \drawPointL[middle][DA,DF]{D}, which is not the centre, straight lines are drawn to the circumference; the greatest is \drawUnitLine{DF,FL}, which passes through the centre: and \drawUnitLine{DC} which is nearer to \drawUnitLine{DF,FL}, $> \drawUnitLine{DB}$ which is more remote \bycref{prop:III.VIII}; but $\drawUnitLine{DC} = \drawUnitLine{DB}$ \bycref{\hypref} which is absurd. +Then since more than two equal straight lines are drawn from a point which is not the centre, to the circumference, two of them at least must lie at the same side of the diameter \drawUnitLine{DH,DF,FL}; and since from a point \drawPointL[middle][DA,DF]{D}, which is not the centre, straight lines are drawn to the circumference; the greatest is \drawUnitLine{DF,FL}, which passes through the centre: and \drawUnitLine{DC} which is nearer to \drawUnitLine{DF,FL}, $> \drawUnitLine{DB}$ which is more remote \byref{prop:III.VIII}; but $\drawUnitLine{DC} = \drawUnitLine{DB}$ \byref{\hypref} which is absurd. The same may be demonstrated of any other point, different from \drawPointL[middle][DA,DF]{D}, which must be the centre of the circle. @@ -5349,9 +5349,9 @@ \chapter*{Definitions} For, if it be possible, let it intersect in three points; from the centre of \drawCircle{PII} draw \drawUnitLine{PG}, \drawUnitLine{PB} and \drawUnitLine{PH} to the points of intersection; -$\therefore \drawUnitLine{PG} = \drawUnitLine{PB} = \drawUnitLine{PH}$ \bycref{def:I.XV}, but as the circles intersect, they have not the same centre \bycref{prop:III.V}: +$\therefore \drawUnitLine{PG} = \drawUnitLine{PB} = \drawUnitLine{PH}$ \byref{def:I.XV}, but as the circles intersect, they have not the same centre \byref{prop:III.V}: -$\therefore$ the assumed point is not the centre of \circlePI , and $\therefore$ as \drawUnitLine{PG}, \drawUnitLine{PB} and \drawUnitLine{PH} are drawn from a point not the centre, they are not equal \bycref{prop:III.VII,prop:III.VIII}; but it was shown before that they were equal, which is absurd; the circles therefore do not intersect in three points. +$\therefore$ the assumed point is not the centre of \circlePI , and $\therefore$ as \drawUnitLine{PG}, \drawUnitLine{PB} and \drawUnitLine{PH} are drawn from a point not the centre, they are not equal \byref{prop:III.VII,prop:III.VIII}; but it was shown before that they were equal, which is absurd; the circles therefore do not intersect in three points. \qed @@ -5401,7 +5401,7 @@ \chapter*{Definitions} stopAutoLabeling; draw byNamedLineFull(A, A, 1, 1, 0, 0)(GF); }; -$\drawSizedLine{GF} + \drawSizedLine{AG} > \drawSizedLine{AF}$ \bycref{prop:I.XX},\\ +$\drawSizedLine{GF} + \drawSizedLine{AG} > \drawSizedLine{AF}$ \byref{prop:I.XX},\\ and $\drawSizedLine{AF} = \drawSizedLine{HD,DG,GF}$ as they are radii of \circleM,\\ but $\drawSizedLine{GF} + \drawSizedLine{AG} > \drawSizedLine{HD,DG,GF}$;\\ take away \drawSizedLine{GF} which is common,\\ @@ -5453,9 +5453,9 @@ \chapter*{Definitions} If it be possible, let \drawUnitLine{FC,CD,DG} joining the centres, and not pass through a point of contact; then from a point of contact draw \drawUnitLine{AF} and \drawUnitLine{AG} to the centres. \begin{center} -$\because \drawUnitLine{AF} + \drawUnitLine{AG} > \drawUnitLine{FC,CD,DG}$ \bycref{prop:I.XX}\\ %because -and $\drawUnitLine{FC} = \drawUnitLine{AF}$ \bycref{def:I.XV},\\ -and $\drawUnitLine{DG} = \drawUnitLine{AG}$ \bycref{def:I.XV}, +$\because \drawUnitLine{AF} + \drawUnitLine{AG} > \drawUnitLine{FC,CD,DG}$ \byref{prop:I.XX}\\ %because +and $\drawUnitLine{FC} = \drawUnitLine{AF}$ \byref{def:I.XV},\\ +and $\drawUnitLine{DG} = \drawUnitLine{AG}$ \byref{def:I.XV}, $\therefore \drawUnitLine{FC} + \drawUnitLine{DG} > \drawUnitLine{FC,CD,DG}$,\\ a part greater than the whole, which is absurd. @@ -5501,19 +5501,19 @@ \chapter*{Definitions} } \startsubproposition{Figure I.}\drawCurrentPictureInMargin -For, if it possible, let \drawCircle{M} and \drawCircle{N} touch one another internally in two points; draw \drawUnitLine{GH} joining their centres, and produce it until it pass through one of the points of contact \bycref{prop:III.XI}; +For, if it possible, let \drawCircle{M} and \drawCircle{N} touch one another internally in two points; draw \drawUnitLine{GH} joining their centres, and produce it until it pass through one of the points of contact \byref{prop:III.XI}; \begin{center} draw \drawUnitLine{DH} and \drawUnitLine{DG}. -But $\drawUnitLine{BG} = \drawUnitLine{DG}$ \bycref{def:I.XV},\\ +But $\drawUnitLine{BG} = \drawUnitLine{DG}$ \byref{def:I.XV},\\ $\therefore$ if \drawUnitLine{GH} be added to both,\\ $\drawUnitLine{GH,BG} = \drawUnitLine{GH} + \drawUnitLine{DG}$; -but $\drawUnitLine{GH,BG} = \drawUnitLine{DH}$ \bycref{def:I.XV},\\ +but $\drawUnitLine{GH,BG} = \drawUnitLine{DH}$ \byref{def:I.XV},\\ and $\therefore \drawUnitLine{GH} + \drawUnitLine{DG} = \drawUnitLine{DH}$; -but $\therefore \drawUnitLine{GH} + \drawUnitLine{DG} > \drawUnitLine{DH}$ \bycref{prop:I.XX},\\ +but $\therefore \drawUnitLine{GH} + \drawUnitLine{DG} > \drawUnitLine{DH}$ \byref{prop:I.XX},\\ which is absurd. \end{center} @@ -5575,12 +5575,12 @@ \chapter*{Definitions} Next, if it be possible, let \drawCircle{H} and \drawCircle{G} touch externally in two points; draw \drawUnitLine{HA,AG} joining the centres of the circles, and passing through one of the points of contact, and draw \drawUnitLine{CH} and \drawUnitLine{CG}. \begin{center} -$\drawUnitLine{CH} = \drawUnitLine{HA}$ \bycref{def:I.XV};\\ -and $\drawUnitLine{AG} = \drawUnitLine{CG}$ \bycref{def:I.XV}; +$\drawUnitLine{CH} = \drawUnitLine{HA}$ \byref{def:I.XV};\\ +and $\drawUnitLine{AG} = \drawUnitLine{CG}$ \byref{def:I.XV}; $\therefore \drawUnitLine{CG} + \drawUnitLine{CH} = \drawUnitLine{HA,AG}$; -but $\drawUnitLine{CG} + \drawUnitLine{CH} > \drawUnitLine{HA,AG}$ \bycref{prop:I.XX}, which is absurd. +but $\drawUnitLine{CG} + \drawUnitLine{CH} > \drawUnitLine{HA,AG}$ \byref{prop:I.XX}, which is absurd. \end{center} There is therefore no case in which two circles can touch one another in two points. @@ -5627,14 +5627,14 @@ \chapter*{Definitions} \begin{center} From the centre of \drawCircle[middle][1/5]{E} draw \drawProportionalLine{EF} $\perp$ to \drawProportionalLine{AF,FB} and $\drawProportionalLine{EG} \perp \drawProportionalLine{CG,GD}$, join \drawProportionalLine{EA} and \drawProportionalLine{EC}. -Then $\drawProportionalLine{CG} = \mbox{ half } \drawProportionalLine{CG,GD}$ \bycref{prop:III.III}\\ -and $\drawProportionalLine{AF} = \frac{1}{2} \drawProportionalLine{AF,FB}$ \bycref{prop:III.III},\\ -since $\drawProportionalLine{CG,GD} = \drawProportionalLine{AF,FB}$ \bycref{\hypref}\\ +Then $\drawProportionalLine{CG} = \mbox{ half } \drawProportionalLine{CG,GD}$ \byref{prop:III.III}\\ +and $\drawProportionalLine{AF} = \frac{1}{2} \drawProportionalLine{AF,FB}$ \byref{prop:III.III},\\ +since $\drawProportionalLine{CG,GD} = \drawProportionalLine{AF,FB}$ \byref{\hypref}\\ $\therefore \drawProportionalLine{CG} = \drawProportionalLine{AF}$,\\ -and $\drawProportionalLine{EA} = \drawProportionalLine{EC}$ \bycref{def:I.XV}\\ +and $\drawProportionalLine{EA} = \drawProportionalLine{EC}$ \byref{def:I.XV}\\ $\therefore \drawProportionalLine{EA}^2 = \drawProportionalLine{EC}^2$;\\ but since \drawAngle{F} is a right angle\\ -$\drawProportionalLine{EA}^2 = \drawProportionalLine{EF}^2 + \drawProportionalLine{AF}^2$ \bycref{prop:I.XLVII}\\ +$\drawProportionalLine{EA}^2 = \drawProportionalLine{EF}^2 + \drawProportionalLine{AF}^2$ \byref{prop:I.XLVII}\\ and $\drawProportionalLine{EC}^2 = \drawProportionalLine{EG}^2 + \drawProportionalLine{CG}^2$ for the same reason,\\ $\therefore \drawProportionalLine{EF}^2 + \drawProportionalLine{AF}^2 = \drawProportionalLine{EG}^2 + \drawProportionalLine{CG}^2$\\ $\therefore \drawProportionalLine{EF}^2 = \drawProportionalLine{EG}^2$\\ @@ -5694,7 +5694,7 @@ \chapter*{Definitions} Then $\drawUnitLine{EM} = \drawUnitLine{EA}$\\ and $\drawUnitLine{EN} = \drawUnitLine{DE}$,\\ $\therefore \drawUnitLine{EN} + \drawUnitLine{EM} = \drawUnitLine{DE,EA}$\\ -but $\drawUnitLine{EN} + \drawUnitLine{EM} > \drawUnitLine{MN}$ \bycref{prop:I.XX} +but $\drawUnitLine{EN} + \drawUnitLine{EM} > \drawUnitLine{MN}$ \byref{prop:I.XX} $\therefore \drawUnitLine{DE,EA} > \drawUnitLine{MN}$. \end{center} @@ -5730,7 +5730,7 @@ \chapter*{Definitions} $\drawUnitLine{EN} \mbox{ and } \drawUnitLine{EM} = \drawUnitLine{EG} \mbox{ and } \drawUnitLine{EF}$;\\ but $\drawAngle{NEG,GEF,FEM} > \drawAngle{GEF}$, -$\therefore \drawUnitLine{MN} > \drawUnitLine{EF}$ \bycref{prop:I.XXIV} +$\therefore \drawUnitLine{MN} > \drawUnitLine{EF}$ \byref{prop:I.XXIV} \end{center} \vfill\pagebreak @@ -5773,8 +5773,8 @@ \chapter*{Definitions} make $\drawUnitLine{EH} = \drawUnitLine{EL}$,\\ and draw $\drawUnitLine{MN} \perp \drawUnitLine{EL,LK}$. -Since \drawUnitLine{BC} and \drawUnitLine{MN} are equally distant from the centre, $\drawUnitLine{BC} = \drawUnitLine{MN}$ \bycref{prop:III.XIV};\\ -but $\drawUnitLine{MN} > \drawUnitLine{FG}$ \bycref{prop:III.XV}, +Since \drawUnitLine{BC} and \drawUnitLine{MN} are equally distant from the centre, $\drawUnitLine{BC} = \drawUnitLine{MN}$ \byref{prop:III.XIV};\\ +but $\drawUnitLine{MN} > \drawUnitLine{FG}$ \byref{prop:III.XV}, $\therefore \drawUnitLine{BC} > \drawUnitLine{FG}$. \end{center} @@ -5827,9 +5827,9 @@ \chapter*{Definitions} \begin{center} Then, $\because \drawUnitLine{DA} = \drawUnitLine{DC}$,\\ %because -$\drawAngle{CAD} = \drawAngle{C}$ \bycref{prop:I.V},\\ -and $\therefore$ each of these angles is acute \bycref{prop:I.XVII}\\ -but $\drawAngle{CAD} = \drawRightAngle$ \bycref{\hypref}, which is absurd, +$\drawAngle{CAD} = \drawAngle{C}$ \byref{prop:I.V},\\ +and $\therefore$ each of these angles is acute \byref{prop:I.XVII}\\ +but $\drawAngle{CAD} = \drawRightAngle$ \byref{\hypref}, which is absurd, therefore \drawUnitLine{AC} drawn $\perp \drawUnitLine{DA}$ does not meet the circle again. \end{center} @@ -5892,7 +5892,7 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem{T}{o}{draw a tangent to a given circle \drawCircle[middle][1/5]{EI} from a given point, either in or outside of its circumference.} -If a given point be in the circumference, as at \drawPointL[middle][FB]{B}, it is plain that the straight line $\drawUnitLine{AB} \perp \drawUnitLine{BE}$ the radius, will be the required tangent \bycref{prop:III.XVI}. +If a given point be in the circumference, as at \drawPointL[middle][FB]{B}, it is plain that the straight line $\drawUnitLine{AB} \perp \drawUnitLine{BE}$ the radius, will be the required tangent \byref{prop:III.XVI}. But if the given point \drawPointL[middle][FB]{A} be outside of the circumference, @@ -5906,7 +5906,7 @@ \chapter*{Definitions} Then \drawUnitLine{AB} will be the tangent required.\\ For in \drawLine[bottom]{FD,FB,BE,DE} and \drawLine[bottom]{BE,DE,AD,AB} $\drawUnitLine{AD,DE} = \drawUnitLine{FB,BE}$, \drawAngle{E} common,\\ and $\drawUnitLine{DE} = \drawUnitLine{BE}$,\\ -$\therefore \mbox{ \bycref{prop:I.IV} } \drawAngle{B} = \drawAngle{D} = \drawRightAngle$,\\ +$\therefore \mbox{ \byref{prop:I.IV} } \drawAngle{B} = \drawAngle{D} = \drawRightAngle$,\\ $\therefore \drawUnitLine{FD}$ is a tangent to \circleEI. \end{center} @@ -5942,9 +5942,9 @@ \chapter*{Definitions} \begin{center} For, if it be possible, let \drawUnitLine{FB,BG} be $\perp \drawUnitLine{CD}$,\\ -then $\because \drawAngle{G} = \drawRightAngle$, \drawAngle{C} is acute \bycref{prop:I.XVII} %because +then $\because \drawAngle{G} = \drawRightAngle$, \drawAngle{C} is acute \byref{prop:I.XVII} %because -$\therefore \drawUnitLine{FC} > \drawUnitLine{FB,BG}$ \bycref{prop:I.XIX}; +$\therefore \drawUnitLine{FC} > \drawUnitLine{FB,BG}$ \byref{prop:I.XIX}; but $\drawUnitLine{FC} = \drawUnitLine{FB}$,\\ and $\therefore \drawUnitLine{FB} > \drawUnitLine{FB,BG}$, a part greater than the whole, which is absurd. @@ -5983,9 +5983,9 @@ \chapter*{Definitions} For, if it be possible, let the centre be without \drawUnitLine{AC}, and draw \drawUnitLine{CF} from the supposed centre to the point of contact. \begin{center} -$\because \drawUnitLine{CF} \perp \drawUnitLine{CE}$ \bycref{prop:III.XVIII}\\ % Because +$\because \drawUnitLine{CF} \perp \drawUnitLine{CE}$ \byref{prop:III.XVIII}\\ % Because $\therefore \drawAngle{FCE} = \drawRightAngle$, a right angle;\\ -but $\drawAngle{ACF,FCE} = \drawRightAngle$ \bycref{\hypref}, +but $\drawAngle{ACF,FCE} = \drawRightAngle$ \byref{\hypref}, and $\therefore \drawAngle{FCE} = \drawAngle{ACF,FCE}$, a part equal to the whole, which is absurd. \end{center} @@ -6024,10 +6024,10 @@ \chapter*{Definitions} a side of \drawAngle{CAF}. $\because \drawUnitLine{EC} = \drawUnitLine{EA}$,\\ %Because -$\drawAngle{CAF} = \drawAngle{C}$ \bycref{prop:I.V}. +$\drawAngle{CAF} = \drawAngle{C}$ \byref{prop:I.V}. But $\drawAngle{CEF} = \drawAngle{CAF} + \drawAngle{C}$,\\ -or $\drawAngle{CEF} = \mbox{ twice } \drawAngle{CAF}$ \bycref{prop:I.XXXII}. +or $\drawAngle{CEF} = \mbox{ twice } \drawAngle{CAF}$ \byref{prop:I.XXXII}. \end{center} \startsubproposition{Figure II.} @@ -6060,7 +6060,7 @@ \chapter*{Definitions} Let the centre be within \drawAngle{BAF,CAF}, the angle at the circumference; draw \drawUnitLine{AF} from the angular point through the centre of the circle;\\ -then $\drawAngle{B} = \drawAngle{BAF}$, and $\drawAngle{C} = \drawAngle{CAF}$, because of the equality of the sides \bycref{prop:I.V}. +then $\drawAngle{B} = \drawAngle{BAF}$, and $\drawAngle{C} = \drawAngle{CAF}$, because of the equality of the sides \byref{prop:I.V}. Hence $\drawAngle{BAF} + \drawAngle{B} + \drawAngle{CAF} + \drawAngle{C} = \mbox{ twice } \drawAngle{BAF,CAF}$. @@ -6139,7 +6139,7 @@ \chapter*{Definitions} Let the segment be greater than a semicircle, and draw \drawUnitLine{DF} and \drawUnitLine{BF} to the centre. \begin{center} -$\drawAngle{F} = \mbox{ twice } \drawAngle{A} \mbox{ or twice } = \drawAngle{E}$ \bycref{prop:III.XX}; +$\drawAngle{F} = \mbox{ twice } \drawAngle{A} \mbox{ or twice } = \drawAngle{E}$ \byref{prop:III.XX}; $\therefore \drawAngle{A} = \drawAngle{E}$ \end{center} @@ -6223,7 +6223,7 @@ \chapter*{Definitions} and $\drawAngle{DAC} = \drawAngle{DBC}$; add \drawAngle{BCA,ACD} to both.\\ -$\drawAngle{DAC,CAB} + \drawAngle{BCA,ACD} = \drawAngle{BCA,ACD} + \drawAngle{CDB} + \drawAngle{DBC} = \mbox{ two right angles }$ \bycref{prop:I.XXXII}. +$\drawAngle{DAC,CAB} + \drawAngle{BCA,ACD} = \drawAngle{BCA,ACD} + \drawAngle{CDB} + \drawAngle{DBC} = \mbox{ two right angles }$ \byref{prop:I.XXXII}. In like manner it may be shown that,\\ $\drawAngle{CDB,BDA} + \drawAngle{ABD,DBC} = \drawTwoRightAngles$. @@ -6286,9 +6286,9 @@ \chapter*{Definitions} draw \drawUnitLine{BC} and \drawUnitLine{BD}. Because the segments are similar,\\ -$\drawAngle{C} = \drawAngle{D}$ \bycref{def:III.XI}, % improvement: definition 10 in the original seems to be irrelevant +$\drawAngle{C} = \drawAngle{D}$ \byref{def:III.XI}, % improvement: definition 10 in the original seems to be irrelevant -but $\drawAngle{C} > \drawAngle{D}$ \bycref{prop:I.XVI}\\ +but $\drawAngle{C} > \drawAngle{D}$ \byref{prop:I.XVI}\\ which is absurd; therefore no point in either of the segments falls without the other, and therefore the segments coincide. @@ -6356,7 +6356,7 @@ \chapter*{Definitions} \drawUnitLine{CD} must wholly coincide with \drawUnitLine{AB}; \end{center} -\noindent and the similar segments being then upon the same straight line and at the same side of it, must also coincide \bycref{prop:III.XXIII}, and are therefore equal. +\noindent and the similar segments being then upon the same straight line and at the same side of it, must also coincide \byref{prop:III.XXIII}, and are therefore equal. \qed @@ -6394,7 +6394,7 @@ \chapter*{Definitions} where they meet is the centre of the circle. \end{center} -Because \drawUnitLine{BC} terminated in the circle is bisected perpendicularly by \drawUnitLine{EF}, it passes through the centre \bycref{prop:III.I}, likewise \drawUnitLine{DF} passes through the centre, therefore the centre is in the intersection of these perpendiculars. +Because \drawUnitLine{BC} terminated in the circle is bisected perpendicularly by \drawUnitLine{EF}, it passes through the centre \byref{prop:III.I}, likewise \drawUnitLine{DF} passes through the centre, therefore the centre is in the intersection of these perpendiculars. \qed @@ -6461,8 +6461,8 @@ \chapter*{Definitions} have\\ $\drawUnitLine{BG} = \drawUnitLine{CG} = \drawUnitLine{EH} = \drawUnitLine{FH}$,\\ and $\drawAngle{G} = \drawAngle{H}$,\\ -$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$ \bycref{prop:I.IV}.\\ -But $\drawAngle{A} = \drawAngle{D}$ \bycref{prop:III.XX};\\ +$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$ \byref{prop:I.IV}.\\ +But $\drawAngle{A} = \drawAngle{D}$ \byref{prop:III.XX};\\ $\therefore$ \drawFromCurrentPicture{ startTempScale(1/5); @@ -6479,8 +6479,8 @@ \chapter*{Definitions} draw byLabelsOnCircle(E, F)(H); stopTempScale; } -are similar \bycref{def:III.XI};\\ % improvement: definition 10 in the original seems to be irrelevant -they are also equal \bycref{prop:III.XXIV} +are similar \byref{def:III.XI};\\ % improvement: definition 10 in the original seems to be irrelevant +they are also equal \byref{prop:III.XXIV} \end{center} If therefore the equal segments be taken from the equal circles, the remaining segments will be equal; @@ -6496,7 +6496,7 @@ \chapter*{Definitions} draw byNamedFilledCircleSegment(H); draw byNamedArcLabel(Hb); } -$ \bycref{ax:I.III};\\ +$ \byref{ax:I.III};\\ and $\therefore \drawArc{Gb} = \drawArc{Hb}$. \end{center} @@ -6567,9 +6567,9 @@ \chapter*{Definitions} \drawAngle{D} be greater than the other \drawAngle{BAK}\\ and make $\drawAngle{BAK,KAC} = \drawAngle{D}$ -$\therefore \drawArc{GbI,GbII} = \drawArc{Hb}$ \bycref{prop:III.XXVI} +$\therefore \drawArc{GbI,GbII} = \drawArc{Hb}$ \byref{prop:III.XXVI} -but $\drawArc{GbI} = \drawArc{Hb}$ \bycref{\hypref}\\ +but $\drawArc{GbI} = \drawArc{Hb}$ \byref{\hypref}\\ $\therefore \drawArc{GbI} = \drawArc{GbI,GbII}$ a part equal to the whole, which is absurd; $\therefore$ neither angle is greater than the other, @@ -6628,13 +6628,13 @@ \chapter*{Definitions} and $\because \circleK = \circleL$\\ % because $\drawUnitLine{AK}, \drawUnitLine{BK} = \drawUnitLine{DL}, \drawUnitLine{EL}$\\ -also $\drawUnitLine{AB} = \drawUnitLine{DE}$ \bycref{\hypref} +also $\drawUnitLine{AB} = \drawUnitLine{DE}$ \byref{\hypref} $\therefore \drawAngle{K} = \drawAngle{L}$ -$\therefore \drawArc{Kb} = \drawArc{Lb}$ \bycref{prop:III.XXVI} +$\therefore \drawArc{Kb} = \drawArc{Lb}$ \byref{prop:III.XXVI} -and $\therefore \drawArc[bottom][1/3]{K} = \drawArc[bottom][1/3]{L}$ \bycref{ax:I.III}. +and $\therefore \drawArc[bottom][1/3]{K} = \drawArc[bottom][1/3]{L}$ \byref{ax:I.III}. \end{center} \qed @@ -6687,12 +6687,12 @@ \chapter*{Definitions} let \drawUnitLine{AK}, \drawUnitLine{BK} and \drawUnitLine{DL}, \drawUnitLine{EL} be drawn to the centres; -$\because \drawArc{Kb} = \drawArc{Lb}$ \bycref{\hypref}\\ % because -and $\drawAngle{K} = \drawAngle{L}$ \bycref{prop:III.XXVII}; +$\because \drawArc{Kb} = \drawArc{Lb}$ \byref{\hypref}\\ % because +and $\drawAngle{K} = \drawAngle{L}$ \byref{prop:III.XXVII}; but $\drawUnitLine{AK} \mbox{ and } \drawUnitLine{BK} = \drawUnitLine{DL} \mbox{ and } \drawUnitLine{EL}$ -$\therefore \drawUnitLine{AB} = \drawUnitLine{DE}$ \bycref{prop:I.IV}; +$\therefore \drawUnitLine{AB} = \drawUnitLine{DE}$ \byref{prop:I.IV}; but these are the chords subtending the equal arcs. \end{center} @@ -6739,11 +6739,11 @@ \chapter*{Definitions} Draw \drawUnitLine{DA} and \drawUnitLine{DB}. -$\drawUnitLine{AC} = \drawUnitLine{CB}$ \bycref{\constref}\\ +$\drawUnitLine{AC} = \drawUnitLine{CB}$ \byref{\constref}\\ \drawUnitLine{DC} is common,\\ % common to what? most likely triangles ACD and CBD -and $\drawAngle{ACD} = \drawAngle{DCB}$ \bycref{\constref} +and $\drawAngle{ACD} = \drawAngle{DCB}$ \byref{\constref} -$\therefore \drawUnitLine{DA} = \drawUnitLine{DB}$ \bycref{prop:I.IV}\\ +$\therefore \drawUnitLine{DA} = \drawUnitLine{DB}$ \byref{prop:I.IV}\\ $ \drawFromCurrentPicture{ startGlobalRotation(-arcAngle.El); @@ -6760,7 +6760,7 @@ \chapter*{Definitions} stopAutoLabeling; stopGlobalRotation; } -$ \bycref{prop:III.XXVIII},\\ +$ \byref{prop:III.XXVIII},\\ and therefore the given arc is bisected. \end{center} @@ -6797,8 +6797,8 @@ \chapter*{Definitions} The angle \drawAngle{EAB,CAE} in a semicircle is a right angle. Draw \drawUnitLine{AE} and \drawUnitLine{BE,EC}\\ -$\drawAngle{B}=\drawAngle{EAB}$ and $\drawAngle{C} = \drawAngle{CAE}$ \bycref{prop:I.V}\\ -$\drawAngle{C} + \drawAngle{B} = \drawAngle{EAB,CAE} = \mbox{ the half of } \drawTwoRightAngles= \drawRightAngle$ \bycref{prop:I.XXXII} +$\drawAngle{B}=\drawAngle{EAB}$ and $\drawAngle{C} = \drawAngle{CAE}$ \byref{prop:I.V}\\ +$\drawAngle{C} + \drawAngle{B} = \drawAngle{EAB,CAE} = \mbox{ the half of } \drawTwoRightAngles= \drawRightAngle$ \byref{prop:I.XXXII} \end{center} \defineNewPicture{ @@ -6855,7 +6855,7 @@ \chapter*{Definitions} The angle \drawAngle{D} in a segment less than a semicircle is obtuse. Take in the opposite circumference any point, to which draw \drawUnitLine{BC} and \drawUnitLine{AB}.\\ -$\because \drawAngle{B} + \drawAngle{D} = \drawTwoRightAngles$ \bycref{prop:III.XXII}\\ % Because +$\because \drawAngle{B} + \drawAngle{D} = \drawTwoRightAngles$ \byref{prop:III.XXII}\\ % Because but $\drawAngle{B} < \drawRightAngle$ (figure II.), $\therefore$ \drawAngle{D} is obtuse. \end{center} @@ -6894,20 +6894,20 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem{I}{f}{a right line \drawUnitLine{EF} be a tangent to a circle, and from the point of contact a right line \drawUnitLine{DB} be drawn cutting the circle, the angle \drawAngle{FBD} made by this line with the tangent is equal to the angle \drawAngle{A} in the alternate segment of the circle.} -If the chord should pass through the centre, it is evident the angles are equal, for each of them is a right angle. \bycref{prop:III.XVI,prop:III.XXXI} +If the chord should pass through the centre, it is evident the angles are equal, for each of them is a right angle. \byref{prop:III.XVI,prop:III.XXXI} -But if not, draw $\drawUnitLine{AB} \perp \drawUnitLine{EF}$ from the point of contact, it must pass through the centre of the circle, \bycref{prop:III.XIX} +But if not, draw $\drawUnitLine{AB} \perp \drawUnitLine{EF}$ from the point of contact, it must pass through the centre of the circle, \byref{prop:III.XIX} \begin{center} -$\therefore \drawAngle{ADB} = \drawAngle{ABE} $ \bycref{prop:III.XXXI} +$\therefore \drawAngle{ADB} = \drawAngle{ABE} $ \byref{prop:III.XXXI} -$\drawAngle{A} + \drawAngle{DBA} = \drawAngle{ABE} = \drawAngle{FBA}$ \bycref{prop:I.XXXII} +$\drawAngle{A} + \drawAngle{DBA} = \drawAngle{ABE} = \drawAngle{FBA}$ \byref{prop:I.XXXII} -$\therefore \drawAngle{A} = \drawAngle{FBD}$ \bycref{ax:I.III}. % improvement: it seems to be axiom III here and down there +$\therefore \drawAngle{A} = \drawAngle{FBD}$ \byref{ax:I.III}. % improvement: it seems to be axiom III here and down there -Again $\drawAngle{B} = \drawTwoRightAngles = \drawAngle{A} + \drawAngle{C}$ \bycref{prop:III.XXII} +Again $\drawAngle{B} = \drawTwoRightAngles = \drawAngle{A} + \drawAngle{C}$ \byref{prop:III.XXII} -$\therefore \drawAngle{DBE} = \drawAngle{C}$, \bycref{ax:I.III}, which is the angle in the alternate segment. +$\therefore \drawAngle{DBE} = \drawAngle{C}$, \byref{ax:I.III}, which is the angle in the alternate segment. \end{center} \qed @@ -6948,7 +6948,7 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem{O}{n}{a given straight line \drawUnitLine{AB} to describe a segment of a circle that shall contain an angle equal to a given angle \drawAngle{givenRight}, \drawAngle{givenObtuse}, \drawAngle{givenAcute}.} -If a given angle be a right angle, bisect the given line, and describe a semicircle on it, this will evidently contain a right angle. \bycref{prop:III.XXXI} +If a given angle be a right angle, bisect the given line, and describe a semicircle on it, this will evidently contain a right angle. \byref{prop:III.XXXI} If the given angle be acute or obtuse, make with the given line, at its extremity, @@ -6960,9 +6960,9 @@ \chapter*{Definitions} describe \drawCircle[middle][1/5]{G} with \drawUnitLine{AG} or \drawUnitLine{GB} as radius, for they are equal. -\drawUnitLine{KD} is tangent to \circleG\ \bycref{prop:III.XVI} +\drawUnitLine{KD} is tangent to \circleG\ \byref{prop:III.XVI} -$\therefore$ \drawUnitLine{AB} divides the circle into two segments capable of containing angles equal to \drawAngle{EAK,BAE} and \drawAngle{DAB} which were made respectively equal to \drawAngle{givenObtuse} and \drawAngle{givenAcute} \bycref{prop:III.XXXII}. +$\therefore$ \drawUnitLine{AB} divides the circle into two segments capable of containing angles equal to \drawAngle{EAK,BAE} and \drawAngle{DAB} which were made respectively equal to \drawAngle{givenObtuse} and \drawAngle{givenAcute} \byref{prop:III.XXXII}. \end{center} \qed @@ -6994,7 +6994,7 @@ \chapter*{Definitions} \problem{T}{o}{cut off from a given circle \drawCircle{O} a segment which shall contain an angle equal to a given angle \drawAngle{givenAngle}.} \begin{center} -Draw \drawUnitLine{EF} \bycref{prop:III.XVII}, a~tangent to the circle at any point; +Draw \drawUnitLine{EF} \byref{prop:III.XVII}, a~tangent to the circle at any point; at the point of contact make $\drawAngle{B} = \drawAngle{givenAngle}$ the given angle; @@ -7005,9 +7005,9 @@ \chapter*{Definitions} Because \drawUnitLine{EF} is a tangent,\\ and \drawUnitLine{AB} cuts it,\\ -the angle in $\segmentO\ = \drawAngle{B}$ \bycref{prop:III.XXXII}, +the angle in $\segmentO\ = \drawAngle{B}$ \byref{prop:III.XXXII}, -but $\drawAngle{B} = \drawAngle{givenAngle}$ \bycref{\constref}. +but $\drawAngle{B} = \drawAngle{givenAngle}$ \byref{\constref}. \end{center} \qed @@ -7068,9 +7068,9 @@ \chapter*{Definitions} \begin{center} Let \drawSizedLine{HC,EH,AE} pass through the centre, and \drawSizedLine{BH,HD} not; draw \drawSizedLine{EB} and \drawSizedLine{DE}. -Then $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EB}^2 - \drawSizedLine{EH}^2$ \bycref{prop:II.VI}, or $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{HC,EH}^2 - \drawSizedLine{EH}^2$. +Then $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EB}^2 - \drawSizedLine{EH}^2$ \byref{prop:II.VI}, or $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{HC,EH}^2 - \drawSizedLine{EH}^2$. -$\therefore \drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EH,AE} \times \drawSizedLine{AE}$ \bycref{prop:II.V}. +$\therefore \drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EH,AE} \times \drawSizedLine{AE}$ \byref{prop:II.V}. \end{center} \defineNewPicture{ @@ -7143,11 +7143,11 @@ \chapter*{Definitions} draw \drawSizedLine{BF} from the centre to the point of contact; -$\drawSizedLine{DB}^2 = \drawSizedLine{CF,DC}^2 - \drawSizedLine{BF}^2$ \bycref{prop:I.XLVII}, %\mbox{ minus } +$\drawSizedLine{DB}^2 = \drawSizedLine{CF,DC}^2 - \drawSizedLine{BF}^2$ \byref{prop:I.XLVII}, %\mbox{ minus } or $\drawSizedLine{DB}^2 = \drawSizedLine{CF,DC}^2 - \drawSizedLine{CF}^2$, %\mbox{ minus } -$\therefore \drawSizedLine{DB}^2 = \drawSizedLine{FA,CF,DC} \times \drawSizedLine{DC}$ \bycref{prop:II.VI}. +$\therefore \drawSizedLine{DB}^2 = \drawSizedLine{FA,CF,DC} \times \drawSizedLine{DC}$ \byref{prop:II.VI}. \end{center} \defineNewPicture{ @@ -7182,11 +7182,11 @@ \chapter*{Definitions} draw \drawSizedLine{EA} and \drawSizedLine{EC}. -Then $\drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DE}^2 - \drawSizedLine{EC}^2$ \bycref{prop:II.VI}, %\mbox{ minus } +Then $\drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DE}^2 - \drawSizedLine{EC}^2$ \byref{prop:II.VI}, %\mbox{ minus } that is, $\drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DE}^2 - \drawSizedLine{EB}^2$, %\mbox{ minus } -$\therefore \drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DB}^2$ \bycref{prop:III.XVIII}. +$\therefore \drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DB}^2$ \byref{prop:III.XVIII}. \end{center} \qed @@ -7228,19 +7228,19 @@ \chapter*{Definitions} \begin{center} Draw from the given point \drawSizedLine{DE}, a tangent to the circle,\\ and draw from the centre \drawSizedLine{DF}, \drawSizedLine{BF}, and \drawSizedLine{EF},\\ -$\drawSizedLine{DE}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \bycref{prop:III.XXXVI}\\ -but $\drawSizedLine{DB}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \bycref{\hypref},\\ +$\drawSizedLine{DE}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \byref{prop:III.XXXVI}\\ +but $\drawSizedLine{DB}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \byref{\hypref},\\ $\therefore \drawSizedLine{DB}^2 = \drawSizedLine{DE}^2$,\\ and $\therefore \drawSizedLine{DB} = \drawSizedLine{DE}$. Then in \drawLine{DB,DF,BF} and \drawLine{EF,DF,DE}\\ $\drawSizedLine{BF} \mbox{ and } \drawSizedLine{DB} = \drawSizedLine{EF} \mbox{ and } \drawSizedLine{DE}$,\\ and \drawSizedLine{DF} is common,\\ -$\therefore \drawAngle{B} = \drawAngle{E}$ \bycref{prop:I.VIII};\\ -but $\drawAngle{E} = \drawRightAngle$ a right angle \bycref{prop:III.XVIII}, +$\therefore \drawAngle{B} = \drawAngle{E}$ \byref{prop:I.VIII};\\ +but $\drawAngle{E} = \drawRightAngle$ a right angle \byref{prop:III.XVIII}, $\therefore \drawAngle{B} = \drawRightAngle$ a right angle,\\ -and $\therefore$ \drawSizedLine{DB} is a tangent to the circle \bycref{prop:III.XVI}. +and $\therefore$ \drawSizedLine{DB} is a tangent to the circle \byref{prop:III.XVI}. \end{center} \qed @@ -7415,14 +7415,14 @@ \chapter*{Definitions} and if $\drawUnitLine{EC,BE} = \drawUnitLine{GH}$, then the problem is solved. But if \drawUnitLine{EC,BE} be not equal to \drawUnitLine{GH},\\ -$\drawUnitLine{EC,BE} > \drawUnitLine{GH}$ \bycref{\hypref}; +$\drawUnitLine{EC,BE} > \drawUnitLine{GH}$ \byref{\hypref}; -make $\drawUnitLine{EC} = \drawUnitLine{GH}$ \bycref{prop:I.III}\\ +make $\drawUnitLine{EC} = \drawUnitLine{GH}$ \byref{prop:I.III}\\ with \drawUnitLine{EC} as radius,\\ describe \drawCircle{C}, cutting \circleO, \\ and draw \drawUnitLine{AC}, which is the line required. -For $\drawUnitLine{AC} = \drawUnitLine{EC} = \drawUnitLine{GH}$ \bycref{prop:I.XV,\constref}. +For $\drawUnitLine{AC} = \drawUnitLine{EC} = \drawUnitLine{GH}$ \byref{prop:I.XV,\constref}. \end{center} \qed @@ -7469,20 +7469,20 @@ \chapter*{Definitions} \problem{I}{n}{a given circle \drawCircle{O} to inscribe a triangle equiangular to a given triangle.} \begin{center} -To any point of the given circle draw \drawUnitLine{GH}, a tangent \bycref{prop:III.XVII}; +To any point of the given circle draw \drawUnitLine{GH}, a tangent \byref{prop:III.XVII}; -and at the point of contact make $\drawAngle{CAH} = \drawAngle{E}$ \bycref{prop:I.XXIII} +and at the point of contact make $\drawAngle{CAH} = \drawAngle{E}$ \byref{prop:I.XXIII} and in like manner $\drawAngle{GAB} = \drawAngle{F}$, and draw \drawUnitLine{BC}. -$\because \drawAngle{CAH} = \drawAngle{E}$ \bycref{\constref}\\ %Because -and $\drawAngle{CAH} = \drawAngle{B}$ \bycref{prop:III.XXXII}\\ +$\because \drawAngle{CAH} = \drawAngle{E}$ \byref{\constref}\\ %Because +and $\drawAngle{CAH} = \drawAngle{B}$ \byref{prop:III.XXXII}\\ $\therefore \drawAngle{B} = \drawAngle{E}$;\\ also $\drawAngle{C} = \drawAngle{F}$ for the same reason. -$\therefore \drawAngle{BAC} = \drawAngle{D}$ \bycref{prop:I.XXXII}, +$\therefore \drawAngle{BAC} = \drawAngle{D}$ \byref{prop:I.XXXII}, \end{center} \noindent and therefore the triangle inscribed in the circle is equiangular to the given one. @@ -7556,10 +7556,10 @@ \chapter*{Definitions} from the centre of the given circle draw \drawUnitLine{KC}, any radius. -Make $\drawAngle{CKA} = \drawAngle{DFH}$ \bycref{prop:I.XXIII}\\ +Make $\drawAngle{CKA} = \drawAngle{DFH}$ \byref{prop:I.XXIII}\\ and $\drawAngle{BKC} = \drawAngle{GED}$. -At the extremities of the radii, draw \drawUnitLine{NL}, \drawUnitLine{LM} and \drawUnitLine{MN}, tangents to the given circle. \bycref{prop:III.XVII} +At the extremities of the radii, draw \drawUnitLine{NL}, \drawUnitLine{LM} and \drawUnitLine{MN}, tangents to the given circle. \byref{prop:III.XVII} The four angles of \drawFromCurrentPicture[bottom]{ @@ -7569,19 +7569,19 @@ \chapter*{Definitions} draw byLabelsOnPolygon(L, A, K, C)(ALL_LABELS, 0); stopTempAngleScale; }, -taken together are equal to four right angles. \bycref{prop:I.XXXII} +taken together are equal to four right angles. \byref{prop:I.XXXII} -but \drawAngle{C} and \drawAngle{A} are right angles \bycref{\constref} +but \drawAngle{C} and \drawAngle{A} are right angles \byref{\constref} $\therefore \drawAngle{L} + \drawAngle{CKA} = \drawTwoRightAngles$, two right angles -but $\drawAngle{DFH,DFE} = \drawTwoRightAngles$ \bycref{prop:I.XIII}\\ -and $\drawAngle{CKA} = \drawAngle{DFH}$ \bycref{\constref} and $\therefore \drawAngle{L} = \drawAngle{DFE}$. +but $\drawAngle{DFH,DFE} = \drawTwoRightAngles$ \byref{prop:I.XIII}\\ +and $\drawAngle{CKA} = \drawAngle{DFH}$ \byref{\constref} and $\therefore \drawAngle{L} = \drawAngle{DFE}$. In the same manner it can be demonstrated that\\ $\drawAngle{N} = \drawAngle{FED}$; -$\therefore \drawAngle{M} = \drawAngle{D}$ \bycref{prop:I.XXXII}\\ +$\therefore \drawAngle{M} = \drawAngle{D}$ \byref{prop:I.XXXII}\\ and therefore the triangle circumscribed about the given circle is equiangular to the given triangle. \end{center} @@ -7632,7 +7632,7 @@ \chapter*{Definitions} stopTempScale; } to inscribe a circle.} -Bisect \drawAngle{EBD,DBF} and \drawAngle{GCD,DCF} \bycref{prop:I.IX} by \drawUnitLine{BD} and \drawUnitLine{CD}; from the point where these lines meet draw \drawUnitLine{FD}, \drawUnitLine{ED} and \drawUnitLine{GD} respectively perpendicular to \drawUnitLine{BC}, \drawUnitLine{AB} and \drawUnitLine{CA}. +Bisect \drawAngle{EBD,DBF} and \drawAngle{GCD,DCF} \byref{prop:I.IX} by \drawUnitLine{BD} and \drawUnitLine{CD}; from the point where these lines meet draw \drawUnitLine{FD}, \drawUnitLine{ED} and \drawUnitLine{GD} respectively perpendicular to \drawUnitLine{BC}, \drawUnitLine{AB} and \drawUnitLine{CA}. \begin{center} In @@ -7654,14 +7654,14 @@ \chapter*{Definitions} stopTempScale; }\\ $\drawAngle{DBF} = \drawAngle{EBD}$, $\drawAngle{F} = \drawAngle{E}$ and \drawUnitLine{BD} common,\\ -$\therefore \drawUnitLine{ED} = \drawUnitLine{FD}$ \bycref{prop:I.IV,prop:I.XXVI}. +$\therefore \drawUnitLine{ED} = \drawUnitLine{FD}$ \byref{prop:I.IV,prop:I.XXVI}. In like manner, it may be shown also that $\drawUnitLine{GD} = \drawUnitLine{FD}$, $\therefore \drawUnitLine{FD} = \drawUnitLine{ED} = \drawUnitLine{GD}$; \end{center} -hence with any one of these lines as radius, describe \drawCircle[middle][1/2]{D} and it will pass through the extremities of the other two; and the sides of the given triangle, being perpendicular to the three radii at their extremities, touch the circle \bycref{prop:III.XVI}, which is therefore inscribed in the given triangle. +hence with any one of these lines as radius, describe \drawCircle[middle][1/2]{D} and it will pass through the extremities of the other two; and the sides of the given triangle, being perpendicular to the three radii at their extremities, touch the circle \byref{prop:III.XVI}, which is therefore inscribed in the given triangle. \qed @@ -7710,9 +7710,9 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem{T}{o}{describe a circle about a given triangle.} -Make $\drawUnitLine{AD} = \drawUnitLine{DB}$ and $\drawUnitLine{CE} = \drawUnitLine{EA}$ \bycref{prop:I.X} +Make $\drawUnitLine{AD} = \drawUnitLine{DB}$ and $\drawUnitLine{CE} = \drawUnitLine{EA}$ \byref{prop:I.X} -From the points of bisection draw \drawUnitLine{DF} and \drawUnitLine{EF} $\perp$ \drawUnitLine{AD} and \drawUnitLine{CE} respectively \bycref{prop:I.XI}, and from their point of concourse draw \drawUnitLine{BF}, \drawUnitLine{AF} and \drawUnitLine{CF} and describe a circle with any one of them, and it will be the circle required. +From the points of bisection draw \drawUnitLine{DF} and \drawUnitLine{EF} $\perp$ \drawUnitLine{AD} and \drawUnitLine{CE} respectively \byref{prop:I.XI}, and from their point of concourse draw \drawUnitLine{BF}, \drawUnitLine{AF} and \drawUnitLine{CF} and describe a circle with any one of them, and it will be the circle required. \begin{center} In @@ -7729,11 +7729,11 @@ \chapter*{Definitions} draw byNamedLineSeq(0)(AF,DF,AD); stopAutoLabeling; }\\ -$\drawUnitLine{DB} = \drawUnitLine{AD}$ \bycref{\constref},\\ +$\drawUnitLine{DB} = \drawUnitLine{AD}$ \byref{\constref},\\ \drawUnitLine{DF} common,\\ -$\drawAngle{FDB} = \drawAngle{ADF}$ \bycref{\constref}, +$\drawAngle{FDB} = \drawAngle{ADF}$ \byref{\constref}, -$\therefore \drawUnitLine{AF} = \drawUnitLine{BF}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{AF} = \drawUnitLine{BF}$ \byref{prop:I.IV}. In like manner it may be shown that $\drawUnitLine{CF} = \drawUnitLine{AF}$. \end{center} @@ -7779,20 +7779,20 @@ \chapter*{Definitions} \begin{center} \drawLine[middle][squareABCD]{DA,CD,BC,AB} is a square. -For, since \drawAngle{A} and \drawAngle{D} are, each of them, in a semicircle, they are right angles \bycref{prop:III.XXXI}, +For, since \drawAngle{A} and \drawAngle{D} are, each of them, in a semicircle, they are right angles \byref{prop:III.XXXI}, -$\therefore \drawUnitLine{CD} \parallel \drawUnitLine{AB}$ \bycref{prop:I.XXVIII}: +$\therefore \drawUnitLine{CD} \parallel \drawUnitLine{AB}$ \byref{prop:I.XXVIII}: and in like manner $\drawUnitLine{DA} \parallel \drawUnitLine{BC}$. -And $\because \drawAngle{AEB} = \drawAngle{DEA}$ \bycref{\constref}, %because +And $\because \drawAngle{AEB} = \drawAngle{DEA}$ \byref{\constref}, %because -and $\drawUnitLine{BE} = \drawUnitLine{AE} = \drawUnitLine{DE}$ \bycref{def:I.XV}. +and $\drawUnitLine{BE} = \drawUnitLine{AE} = \drawUnitLine{DE}$ \byref{def:I.XV}. -$\therefore \drawUnitLine{AB} = \drawUnitLine{DA}$ \bycref{prop:I.IV}; +$\therefore \drawUnitLine{AB} = \drawUnitLine{DA}$ \byref{prop:I.IV}; \end{center} -and since the adjacent sides and angles of the parallelogram \squareABCD\ are equal, they are all equal \bycref{prop:I.XXXIV}; and $\therefore$ \squareABCD, inscribed in the given circle, is a square. +and since the adjacent sides and angles of the parallelogram \squareABCD\ are equal, they are all equal \byref{prop:I.XXXIV}; and $\therefore$ \squareABCD, inscribed in the given circle, is a square. \qed @@ -7836,15 +7836,15 @@ \chapter*{Definitions} \begin{center} and \drawLine[bottom][squareFGHK]{KF,HK,GH,FG} is a square. -$\drawAngle{C} = \drawRightAngle$ a right angle, \bycref{prop:III.XVIII} +$\drawAngle{C} = \drawRightAngle$ a right angle, \byref{prop:III.XVIII} -also $\drawAngle{E} = \drawRightAngle$ \bycref{\constref}, +also $\drawAngle{E} = \drawRightAngle$ \byref{\constref}, $\therefore \drawUnitLine{HK} \parallel \drawUnitLine{BD}$; in the same manner it can be demonstrated that $\drawUnitLine{FG} \parallel \drawUnitLine{BD}$, and also that $\drawUnitLine{GH} \mbox{ and } \drawUnitLine{KF} \parallel \drawUnitLine{AC}$; $\therefore$ \squareFGHK\ is a parallelogram, -and $\because \drawAngle{C} = \drawAngle{G} = \drawAngle{F} = \drawAngle{K} = \drawAngle{H}$ they are all right angles \bycref{prop:I.XXXIV}; % because +and $\because \drawAngle{C} = \drawAngle{G} = \drawAngle{F} = \drawAngle{K} = \drawAngle{H}$ they are all right angles \byref{prop:I.XXXIV}; % because it is also evident that \drawUnitLine{HK}, \drawUnitLine{GH}, \drawUnitLine{FG} and \drawUnitLine{KF} are equal. @@ -7892,21 +7892,21 @@ \chapter*{Definitions} and $\drawUnitLine{KC} = \drawUnitLine{DK}$,\\ draw $\drawUnitLine{FG,GK} \parallel \drawUnitLine{AE,ED}$,\\ and $\drawUnitLine{EG,GH} \parallel \drawUnitLine{DK,KC}$\\ -\bycref{prop:I.XXXI} +\byref{prop:I.XXXI} $\therefore$ \drawPolygon[bottom]{EDKG} is a parallelogram; -and since $\drawUnitLine{AE,ED} = \drawUnitLine{DK,KC}$ \bycref{\hypref}\\ +and since $\drawUnitLine{AE,ED} = \drawUnitLine{DK,KC}$ \byref{\hypref}\\ $\drawUnitLine{ED} = \drawUnitLine{DK}$ -$\therefore$ \polygonEDKG\ is equilateral \bycref{prop:I.XXXIV} +$\therefore$ \polygonEDKG\ is equilateral \byref{prop:I.XXXIV} In like manner it can be shown that $ \drawPolygon[bottom]{GKCH} = \drawPolygon[bottom]{FGHB}$ are equilateral parallelograms; $\therefore \drawUnitLine{AE} = \drawUnitLine{GK} = \drawUnitLine{GH} = \drawUnitLine{FG}$. \end{center} -\noindent and therefore if a circle be described from the concourse of these lines with any one of them as radius, it will be inscribed in the given square \bycref{prop:III.XVI}. +\noindent and therefore if a circle be described from the concourse of these lines with any one of them as radius, it will be inscribed in the given square \byref{prop:III.XVI}. \qed @@ -7944,7 +7944,7 @@ \chapter*{Definitions} then, because \drawPolygon{ACD} and \drawPolygon{ABC} have their sides equal, and the base \drawUnitLine{AE,EC} common to both, \begin{center} -$\drawAngle{DAE} = \drawAngle{EAB}$ \bycref{prop:I.VIII},\\ +$\drawAngle{DAE} = \drawAngle{EAB}$ \byref{prop:I.VIII},\\ or \drawAngle{DAE,EAB} is bisected: \\ in like manner it can be shown that \drawAngle{ABE,EBC} is bisected; @@ -7952,7 +7952,7 @@ \chapter*{Definitions} hence $\drawAngle{EAB} = \drawAngle{ABE}$ their halves, -$\therefore \drawUnitLine{BE} = \drawUnitLine{AE}$ \bycref{prop:I.VI}; +$\therefore \drawUnitLine{BE} = \drawUnitLine{AE}$ \byref{prop:I.VI}; and in like manner it can be proved that \\ $\drawUnitLine{AE} = \drawUnitLine{BE} = \drawUnitLine{EC} = \drawUnitLine{ED}$. @@ -8001,27 +8001,27 @@ \chapter*{Definitions} \begin{center} Take any straight line \drawProportionalLine{AC,CB} \\ -and divide it so that $\drawProportionalLine{AC,CB} \times \drawProportionalLine{CB} = \drawProportionalLine{AC}^2$ \bycref{prop:II.XI} +and divide it so that $\drawProportionalLine{AC,CB} \times \drawProportionalLine{CB} = \drawProportionalLine{AC}^2$ \byref{prop:II.XI} -With \drawProportionalLine{AC,CB} as radius, describe \drawCircle[middle][1/5]{A} and place in it from the extremity of the radius, $\drawProportionalLine{BD} = \drawProportionalLine{AC}$ \bycref{prop:IV.I};\\ +With \drawProportionalLine{AC,CB} as radius, describe \drawCircle[middle][1/5]{A} and place in it from the extremity of the radius, $\drawProportionalLine{BD} = \drawProportionalLine{AC}$ \byref{prop:IV.I};\\ draw \drawProportionalLine{AD}. Then \drawLine{AD,BD,CB,AC} is the required triangle. -For, draw \drawProportionalLine{CD} and describe \drawCircle[middle][1/3]{F} about \drawLine{AD,CD,AC} \bycref{prop:IV.V} +For, draw \drawProportionalLine{CD} and describe \drawCircle[middle][1/3]{F} about \drawLine{AD,CD,AC} \byref{prop:IV.V} Since $\drawProportionalLine{AC,CB} \times \drawProportionalLine{CB} = \drawProportionalLine{AC}^2 = \drawProportionalLine{BD}^2$,\\ -$\therefore$ \drawProportionalLine{BD} is tangent to \circleF\ \bycref{prop:III.XXXVII}\\ -$\therefore \drawAngle{BDC} = \drawAngle{A}$ \bycref{prop:III.XXXII},\\ +$\therefore$ \drawProportionalLine{BD} is tangent to \circleF\ \byref{prop:III.XXXVII}\\ +$\therefore \drawAngle{BDC} = \drawAngle{A}$ \byref{prop:III.XXXII},\\ add \drawAngle{CDA} to each,\\ $\therefore \drawAngle{BDC} + \drawAngle{CDA} = \drawAngle{A} + \drawAngle{CDA}$; -but $\drawAngle{BDC} + \drawAngle{CDA} \mbox{ or } \drawAngle{BDC,CDA} = \drawAngle{B}$ \bycref{prop:I.V}:\\ -since $\drawProportionalLine{AD} = \drawProportionalLine{AC,CB}$ \bycref{prop:I.V}\\ -consequently $\drawAngle{B} = \drawAngle{A} + \drawAngle{CDA} = \drawAngle{C}$ \bycref{prop:I.XXXII}\\ -$\therefore \drawProportionalLine{CD} = \drawProportionalLine{BD}$ \bycref{prop:I.VI}\\ -$\therefore \drawProportionalLine{BD} = \drawProportionalLine{AC} = \drawProportionalLine{CD}$ \bycref{\constref}\\ -$\therefore \drawAngle{A} = \drawAngle{CDA}$ \bycref{prop:I.V} +but $\drawAngle{BDC} + \drawAngle{CDA} \mbox{ or } \drawAngle{BDC,CDA} = \drawAngle{B}$ \byref{prop:I.V}:\\ +since $\drawProportionalLine{AD} = \drawProportionalLine{AC,CB}$ \byref{prop:I.V}\\ +consequently $\drawAngle{B} = \drawAngle{A} + \drawAngle{CDA} = \drawAngle{C}$ \byref{prop:I.XXXII}\\ +$\therefore \drawProportionalLine{CD} = \drawProportionalLine{BD}$ \byref{prop:I.VI}\\ +$\therefore \drawProportionalLine{BD} = \drawProportionalLine{AC} = \drawProportionalLine{CD}$ \byref{\constref}\\ +$\therefore \drawAngle{A} = \drawAngle{CDA}$ \byref{prop:I.V} $\therefore \drawAngle{B} = \drawAngle{BDC,CDA} = \drawAngle{C} = \drawAngle{A} + \drawAngle{CDA} = \mbox{ twice } \drawAngle{A}$; @@ -8068,14 +8068,14 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem[2]{I}{n}{a given circle \drawCircle[middle][1/4]{O} to inscribe an equilateral and equiangular pentagon.} -Construct an isosceles triangle, in which each of the angles at the base shall be double of the angle at the vertex, and inscribe in the given circle a triangle \drawPolygon[bottom]{ACD} equiangular to it \bycref{prop:IV.II}; +Construct an isosceles triangle, in which each of the angles at the base shall be double of the angle at the vertex, and inscribe in the given circle a triangle \drawPolygon[bottom]{ACD} equiangular to it \byref{prop:IV.II}; \begin{center} -Bisect \drawAngle{ECA,DCE} and \drawAngle{ADB,BDC} \bycref{prop:I.IX},\\ +Bisect \drawAngle{ECA,DCE} and \drawAngle{ADB,BDC} \byref{prop:I.IX},\\ draw \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{EA} and \drawUnitLine{DE}. \end{center} -Because each of the angles \drawAngle{ECA}, \drawAngle{DCE}, \drawAngle{CAD}, \drawAngle{BDC} and \drawAngle{ADB} are equal, the arcs upon which they stand are equal \bycref{prop:III.XXVI}; and $\therefore$ \drawUnitLine{CD}, \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{EA} and \drawUnitLine{DE} which subtend these arcs are equal \bycref{prop:III.XXIX} and $\therefore$ the pentagon is equilateral, it is also equiangular, as each of its angles stand upon equal arcs \bycref{prop:III.XXVII}. +Because each of the angles \drawAngle{ECA}, \drawAngle{DCE}, \drawAngle{CAD}, \drawAngle{BDC} and \drawAngle{ADB} are equal, the arcs upon which they stand are equal \byref{prop:III.XXVI}; and $\therefore$ \drawUnitLine{CD}, \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{EA} and \drawUnitLine{DE} which subtend these arcs are equal \byref{prop:III.XXIX} and $\therefore$ the pentagon is equilateral, it is also equiangular, as each of its angles stand upon equal arcs \byref{prop:III.XXVII}. \qed @@ -8126,7 +8126,7 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem{T}{o}{describe an equilateral and equiangular pentagon about a given circle \drawCircle[middle][1/4]{F}.} -Draw five tangents through the vertices of the angles of any regular pentagon inscribed in the given circle \circleF\ \bycref{prop:III.XVII}. +Draw five tangents through the vertices of the angles of any regular pentagon inscribed in the given circle \circleF\ \byref{prop:III.XVII}. These five tangents will form the required pentagon. @@ -8141,17 +8141,17 @@ \chapter*{Definitions} }\right.$. In \drawLine{FB,FK,BK} and \drawLine{FK,FC,KC}\\ -$\drawProportionalLine{BK} = \drawProportionalLine{KC}$ \bycref{prop:I.XLVII} \\ +$\drawProportionalLine{BK} = \drawProportionalLine{KC}$ \byref{prop:I.XLVII} \\ $\drawProportionalLine{FC} = \drawProportionalLine{FB}$, and \drawProportionalLine{FK} common; -$\therefore \drawAngle{FKB} = \drawAngle{CKF}$ and $\therefore \drawAngle{BFK} = \drawAngle{KFC}$ \bycref{prop:I.VIII} +$\therefore \drawAngle{FKB} = \drawAngle{CKF}$ and $\therefore \drawAngle{BFK} = \drawAngle{KFC}$ \byref{prop:I.VIII} $\therefore \drawAngle{FKB,CKF} = \mbox{ twice } \drawAngle{CKF}$, and $\drawAngle{BFK,KFC} = \mbox{ twice } \drawAngle{KFC}$. In the same manner it can be demonstrated that\\ $\drawAngle{FLC,DLF} = \mbox{ twice } \drawAngle{FLC}$, and $\drawAngle{CFL,LFD} = \mbox{ twice } \drawAngle{CFL}$; -but $\drawAngle{BFK,KFC} = \drawAngle{CFL,LFD}$ \bycref{prop:III.XXVII}, +but $\drawAngle{BFK,KFC} = \drawAngle{CFL,LFD}$ \byref{prop:III.XXVII}, $\therefore$ their halves $\drawAngle{KFC} = \drawAngle{CFL}$, also $\drawAngle{FCK} = \drawAngle{LCF}$,\\ and \drawProportionalLine{FC} common; @@ -8238,7 +8238,7 @@ \chapter*{Definitions} be a given equi\-an\-gu\-lar and equilateral pentagon; it is required to inscribe a circle in it. \begin{center} -Make $\drawAngle{HCF} = \drawAngle{FCK}$, and $\drawAngle{FDE} = \drawAngle{KDF}$ \bycref{prop:I.IX} +Make $\drawAngle{HCF} = \drawAngle{FCK}$, and $\drawAngle{FDE} = \drawAngle{KDF}$ \byref{prop:I.IX} Draw \drawUnitLine{FD}, \drawUnitLine{FC}, \drawUnitLine{FB}, \drawUnitLine{FE}, \&c. @@ -8262,7 +8262,7 @@ \chapter*{Definitions} stopTempScale; }; -$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ and $\drawAngle{FBH} = \drawAngle{KDF}$ \bycref{prop:I.IV} +$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ and $\drawAngle{FBH} = \drawAngle{KDF}$ \byref{prop:I.IV} And $\because \drawAngle{B} = \drawAngle{D} = \mbox{ twice } \drawAngle{KDF}$ %because @@ -8290,15 +8290,15 @@ \chapter*{Definitions} stopAutoLabeling; stopTempScale; }\\ -we have $\drawAngle{HCF} = \drawAngle{FCK}$ \bycref{\constref}, \drawUnitLine{FC} common,\\ +we have $\drawAngle{HCF} = \drawAngle{FCK}$ \byref{\constref}, \drawUnitLine{FC} common,\\ and $\drawAngle{H} = \drawAngle{K} = \mbox{ a right angle }$; -$\therefore \drawUnitLine{FK} = \drawUnitLine{FH}$ \bycref{prop:I.XXVI} +$\therefore \drawUnitLine{FK} = \drawUnitLine{FH}$ \byref{prop:I.XXVI} \end{center} In the same way it may be shown that the five perpendiculars on the sides of the pentagon are equal to one another. -Describe \drawCircle[middle][1/5]{F} with any one of the perpendiculars as radius, and it will be the inscribed circle required. For if it does not touch the sides of the pentagon, but cut them, then a line drawn from the extremity at right angles to the diameter of a circle will fall within the circle, which has been shown to be absurd \bycref{prop:III.XVI}. +Describe \drawCircle[middle][1/5]{F} with any one of the perpendiculars as radius, and it will be the inscribed circle required. For if it does not touch the sides of the pentagon, but cut them, then a line drawn from the extremity at right angles to the diameter of a circle will fall within the circle, which has been shown to be absurd \byref{prop:III.XVI}. \qed @@ -8345,7 +8345,7 @@ \chapter*{Definitions} $\drawAngle{BCF,FCD} = \drawAngle{CDF,FDE}$,\\ $\drawAngle{FCD} = \drawAngle{CDF}$, -$\therefore \drawUnitLine{FD} = \drawUnitLine{FC}$ \bycref{prop:I.VI}; +$\therefore \drawUnitLine{FD} = \drawUnitLine{FC}$ \byref{prop:I.VI}; and since in \drawFromCurrentPicture{ @@ -8364,7 +8364,7 @@ \chapter*{Definitions} $\drawUnitLine{BC} = \drawUnitLine{CD}$, and \drawUnitLine{FC} common,\\ also $\drawAngle{BCF} = \drawAngle{FCD}$; -$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ \byref{prop:I.IV}. In like manner it may be proved that\\ $\drawUnitLine{FA} = \drawUnitLine{FE} = \drawUnitLine{FB}$. @@ -8436,9 +8436,9 @@ \chapter*{Definitions} draw byNamedLineSeq(0)(GD,GE,DE); stopAutoLabeling; } -are equilateral triangles, hence $\drawAngle{DGC} = \drawAngle{EGD} = \mbox{ one-third of } \drawTwoRightAngles$ \bycref{prop:I.XXXII} but $\drawAngle{DGC,EGD,FGE} = \drawTwoRightAngles$ \bycref{prop:I.XIII}; +are equilateral triangles, hence $\drawAngle{DGC} = \drawAngle{EGD} = \mbox{ one-third of } \drawTwoRightAngles$ \byref{prop:I.XXXII} but $\drawAngle{DGC,EGD,FGE} = \drawTwoRightAngles$ \byref{prop:I.XIII}; -$\therefore \drawAngle{DGC} = \drawAngle{EGD} = \drawAngle{FGE} = \mbox{ one-third of } \drawTwoRightAngles$ \bycref{prop:I.XXXII}, and the angles vertically opposite to these are all equal to one another \bycref{prop:I.XV}, and stand on equal arches \bycref{prop:III.XXVI}, which are subtended by equal chords \bycref{prop:III.XXIX}; and since each of the angles of the hexagon is double of the angle of an equilateral triangle, it is also equiangular. +$\therefore \drawAngle{DGC} = \drawAngle{EGD} = \drawAngle{FGE} = \mbox{ one-third of } \drawTwoRightAngles$ \byref{prop:I.XXXII}, and the angles vertically opposite to these are all equal to one another \byref{prop:I.XV}, and stand on equal arches \byref{prop:III.XXVI}, which are subtended by equal chords \byref{prop:III.XXIX}; and since each of the angles of the hexagon is double of the angle of an equilateral triangle, it is also equiangular. \qed @@ -8488,7 +8488,7 @@ \chapter*{Definitions} $\therefore$ the arc subtended by $\drawUnitLine{CF} = \dfrac{1}{15}$ difference of the whole circumference. -Hence if straight lines equal to \drawUnitLine{CF} be placed in the circle \bycref{prop:IV.I}, an equilateral and equiangular quindecagon will be thus inscribed in the circle. +Hence if straight lines equal to \drawUnitLine{CF} be placed in the circle \byref{prop:IV.I}, an equilateral and equiangular quindecagon will be thus inscribed in the circle. \qed @@ -8844,9 +8844,9 @@ \chapter*{Axioms} } \problem{I}{f}{the first of four magnitudes have the same ratio to the second, which the third has to the fourth, then any equimultiples whatever of the first and third shall have the same ratio to any equimultiples of the second and fourth; viz., the equimultiple of the first shall have the same ratio to that of the second, which the equimultiple of the third has to that of the fourth.} -Let $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$, then $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$, every equimultiple of $3\drawMagnitude{I}$ and $3\drawMagnitude{III}$ are equimultiples of \drawMagnitude{I} and \drawMagnitude{III}, and every equimultiple of $2\drawMagnitude{II}$ and $2\drawMagnitude{IV}$, are equimultiples of \drawMagnitude{II} and \drawMagnitude{IV} \bycref{prop:V.III} +Let $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$, then $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$, every equimultiple of $3\drawMagnitude{I}$ and $3\drawMagnitude{III}$ are equimultiples of \drawMagnitude{I} and \drawMagnitude{III}, and every equimultiple of $2\drawMagnitude{II}$ and $2\drawMagnitude{IV}$, are equimultiples of \drawMagnitude{II} and \drawMagnitude{IV} \byref{prop:V.III} -That is, $M$ times $3\drawMagnitude{I}$ and $M$ times $3\drawMagnitude{III}$ are equimultiples of \drawMagnitude{I} and \drawMagnitude{III}, and $m$ times $2\drawMagnitude{II}$ and $m 2\drawMagnitude{IV}$ are equimultiples of $2\drawMagnitude{II}$ and $2\drawMagnitude{IV}$; but $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$ \bycref{\hypref}; $\therefore$ if $M 3\drawMagnitude{I} <, =, \mbox{ or } > m 2 \drawMagnitude{II}$, then $M 3\drawMagnitude{III} <, =, \mbox{ or } > m 2 \drawMagnitude{IV}$ \bycref{def:V.V} and therefore $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$ \bycref{def:V.V} +That is, $M$ times $3\drawMagnitude{I}$ and $M$ times $3\drawMagnitude{III}$ are equimultiples of \drawMagnitude{I} and \drawMagnitude{III}, and $m$ times $2\drawMagnitude{II}$ and $m 2\drawMagnitude{IV}$ are equimultiples of $2\drawMagnitude{II}$ and $2\drawMagnitude{IV}$; but $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$ \byref{\hypref}; $\therefore$ if $M 3\drawMagnitude{I} <, =, \mbox{ or } > m 2 \drawMagnitude{II}$, then $M 3\drawMagnitude{III} <, =, \mbox{ or } > m 2 \drawMagnitude{IV}$ \byref{def:V.V} and therefore $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$ \byref{def:V.V} The same reasoning holds good if any other equimultiple of the first and third be taken, any other equimultiple of the second and fourth. @@ -9080,9 +9080,9 @@ \chapter*{Axioms} take \magnitudeVI\ the same multiple of \magnitudeIV, \\ then, $\because \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$\\ %because and of the second and fourth, we have taken equimultiples,\\ -\magnitudeI\ and \magnitudeVI\ therefore \bycref{prop:V.IV}\\ - $\magnitudeI : \magnitudeV :: \magnitudeIII : \magnitudeVI$, but \bycref{\constref},\\ - $\magnitudeI = \magnitudeV \therefore$ \bycref{prop:V.A} $\magnitudeIII = \magnitudeVI$\\ +\magnitudeI\ and \magnitudeVI\ therefore \byref{prop:V.IV}\\ + $\magnitudeI : \magnitudeV :: \magnitudeIII : \magnitudeVI$, but \byref{\constref},\\ + $\magnitudeI = \magnitudeV \therefore$ \byref{prop:V.A} $\magnitudeIII = \magnitudeVI$\\ and \magnitudeVI\ is the same multiple of \magnitudeIV\\ that \magnitudeI\ is of \magnitudeII. @@ -9090,7 +9090,7 @@ \chapter*{Axioms} and also \magnitudeII\ a part of \magnitudeI;\\ then \magnitudeIV\ shall be the same part of \magnitudeIII. -Inversely \bycref{prop:V.B}, $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$,\\ % improvement: in the original there was a reference to "B. 5." which doesn't make much sense +Inversely \byref{prop:V.B}, $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$,\\ % improvement: in the original there was a reference to "B. 5." which doesn't make much sense but \magnitudeII\ is a part of \magnitudeI;\\ that is, \magnitudeI\ is a multiple of \magnitudeII;\\ $\therefore$ by the preceding case, \magnitudeIII\ is the same multiple of \magnitudeIV\\ @@ -9119,12 +9119,12 @@ \chapter*{Axioms} $\therefore$ if $M \magnitudeI >, = \mbox{ or } < m \magnitudeIII$, then\\ $M \magnitudeII >, = \mbox{ or } < m \magnitudeIII$,\\ -and $\therefore \magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \bycref{def:V.V}. +and $\therefore \magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \byref{def:V.V}. From the foregoing reasoning it is evident that,\\ if $m \magnitudeIII >, = \mbox{ or } < M \magnitudeI$, then\\ $m \magnitudeIII >, = \mbox{ or } < M \magnitudeII$\\ -$\therefore \magnitudeIII : \magnitudeI = \magnitudeIII : \magnitudeII$ \bycref{def:V.V}. +$\therefore \magnitudeIII : \magnitudeI = \magnitudeIII : \magnitudeII$ \byref{def:V.V}. $\therefore$ Equal magnitudes, \&c. \end{center} @@ -9309,7 +9309,7 @@ \chapter*{Axioms} Let $\drawMagnitude{I} : \drawMagnitude{III} :: \drawMagnitude{II} : \drawMagnitude{III}$, then $\magnitudeI = \magnitudeII$. For, if not, let $\magnitudeI > \magnitudeII$, then will\\ -$\magnitudeI : \magnitudeIII > \magnitudeII : \magnitudeIII$ \bycref{prop:V.VIII}, +$\magnitudeI : \magnitudeIII > \magnitudeII : \magnitudeIII$ \byref{prop:V.VIII}, which is absurd according to the hypothesis. $\therefore \magnitudeI \mbox{ is } \ngtr \magnitudeII$. % \mbox{ is not } > @@ -9348,8 +9348,8 @@ \chapter*{Axioms} Let $\drawMagnitude{I} : \drawMagnitude{III} > \drawMagnitude{II} : \drawMagnitude{III}$, then $\magnitudeI > \magnitudeII$. For if not, let $\magnitudeI = \mbox{ or } < \magnitudeII$;\\ -then, $\magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \bycref{prop:V.VII} or\\ -$\magnitudeI : \magnitudeIII < \magnitudeII : \magnitudeIII$ \bycref{prop:V.VIII} and (invert.), which is absurd according to the hypothesis. +then, $\magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \byref{prop:V.VII} or\\ +$\magnitudeI : \magnitudeIII < \magnitudeII : \magnitudeIII$ \byref{prop:V.VIII} and (invert.), which is absurd according to the hypothesis. $\therefore \magnitudeI \mbox{ is } \neq \mbox{ or } < \magnitudeII$, and\\ % \mbox{ is not } = $\therefore \magnitudeI \mbox{ must be } >\magnitudeII$. @@ -9358,8 +9358,8 @@ \chapter*{Axioms} then, $\magnitudeII < \magnitudeI$. For if not, $\magnitudeII \mbox{ must be } > \mbox{ or } = \magnitudeI$,\\ -then $\magnitudeIII : \magnitudeII < \magnitudeIII : \magnitudeI$ \bycref{prop:V.VIII} and (invert.);\\ -or $\magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeI$ \bycref{prop:V.VII}, which is absurd \bycref{\hypref}; +then $\magnitudeIII : \magnitudeII < \magnitudeIII : \magnitudeI$ \byref{prop:V.VIII} and (invert.);\\ +or $\magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeI$ \byref{prop:V.VII}, which is absurd \byref{\hypref}; $\therefore \magnitudeII \mbox{ is } \ngtr \mbox{ or } = \magnitudeI$,\\ % \mbox{ is not } > and $\therefore \magnitudeII \mbox{ must be } < \magnitudeI$. @@ -9387,10 +9387,10 @@ \chapter*{Axioms} For if $M \magnitudeI >, = \mbox{ or } < m \magnitudeII$,\\ then $M \magnitudeV >, = \mbox{ or } < m \magnitudeVI$,\\ and if $M \magnitudeV >, = \mbox{ or } < m \magnitudeVI$,\\ -then $M \magnitudeIII >, = \mbox{ or } < m \magnitudeIV$ \bycref{def:V.V}; +then $M \magnitudeIII >, = \mbox{ or } < m \magnitudeIV$ \byref{def:V.V}; $\therefore$ if $M \magnitudeI >, = \mbox{ or } < m \magnitudeII$, $M \magnitudeIII >, = \mbox{ or } < m \magnitudeIV$\\ -and $\therefore$ \bycref{def:V.V} $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$. +and $\therefore$ \byref{def:V.V} $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$. $\therefore$ Ratios that are the same \&c. \end{center} @@ -9424,7 +9424,7 @@ \chapter*{Axioms} For if $M \magnitudeI > m \magnitudeII$, then $M \magnitudeIII > m \magnitudeIV$,\\ and $M \magnitudeV > m \magnitudeVI$, $M \magnitudeVII > m \magnitudeVIII$,\\ -also $M \magnitudeIX > m \magnitudeX$ \bycref{def:V.V} +also $M \magnitudeIX > m \magnitudeX$ \byref{def:V.V} Therefore, if $M \magnitudeI > m \magnitudeII$, then will\\ $M \magnitudeI + M \magnitudeIII + M \magnitudeV + M \magnitudeVII + M \magnitudeIX$, or $M (\magnitudeI + \magnitudeIII + \magnitudeV + \magnitudeVII + \magnitudeIX)$ be greater than $m \magnitudeII + m \magnitudeIV + m \magnitudeVI + m \magnitudeVIII + m \magnitudeX$, or $m (\magnitudeII + \magnitudeIV + \magnitudeVI + \magnitudeVIII + \magnitudeX)$. @@ -9457,7 +9457,7 @@ \chapter*{Axioms} Let these multiples be taken, and take the same multiples of \magnitudeI\ and \magnitudeII. -$\therefore$ \bycref{def:V.V} if $M' \magnitudeI >, =, \mbox{ or } < m' \magnitudeII$;\\ +$\therefore$ \byref{def:V.V} if $M' \magnitudeI >, =, \mbox{ or } < m' \magnitudeII$;\\ then will $M' \magnitudeIII >, =, \mbox{ or } < m' \magnitudeIV$,\\ but $M' \magnitudeIII > m' \magnitudeIV$ (construction); @@ -9484,16 +9484,16 @@ \chapter*{Axioms} Let $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$, and first suppose\\ $\magnitudeI > \magnitudeIII$, then will $\magnitudeII > \magnitudeIV$. -For $\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \bycref{prop:V.VIII}, and by the hypothesis $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$;\\ -$\therefore \magnitudeIII : \magnitudeIV > \magnitudeIII : \magnitudeII$ \bycref{prop:V.XIII},\\ -$\therefore \magnitudeIV < \magnitudeII$ \bycref{prop:V.X}, or $\magnitudeII > \magnitudeIV$. +For $\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \byref{prop:V.VIII}, and by the hypothesis $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$;\\ +$\therefore \magnitudeIII : \magnitudeIV > \magnitudeIII : \magnitudeII$ \byref{prop:V.XIII},\\ +$\therefore \magnitudeIV < \magnitudeII$ \byref{prop:V.X}, or $\magnitudeII > \magnitudeIV$. Secondly, let $\magnitudeI = \magnitudeIII$, then will $\magnitudeII = \magnitudeIV$. -For $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \bycref{prop:V.VII},\\ -and $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$ \bycref{\hypref};\\ -$\therefore \magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XI},\\ -and $\therefore \magnitudeII = \magnitudeIV$ \bycref{prop:V.IX}. +For $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \byref{prop:V.VII},\\ +and $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$ \byref{\hypref};\\ +$\therefore \magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeIV$ \byref{prop:V.XI},\\ +and $\therefore \magnitudeII = \magnitudeIV$ \byref{prop:V.IX}. Thirdly, if $\magnitudeI < \magnitudeIII$, then will $\magnitudeII < \magnitudeIV$;\\ $\because \magnitudeIII > \magnitudeI$ and $\magnitudeIII : \magnitudeIV = \magnitudeI : \magnitudeII$;\\ %because @@ -9521,7 +9521,7 @@ \chapter*{Axioms} &= \magnitudeI : \magnitudeII \\ &= \magnitudeI : \magnitudeII \\ \end{aligned}$\\ -$\therefore \magnitudeI : \magnitudeII :: 4 \magnitudeI : 4 \magnitudeII$. \bycref{prop:V.XII}. +$\therefore \magnitudeI : \magnitudeII :: 4 \magnitudeI : 4 \magnitudeII$. \byref{prop:V.XII}. And as the same reasoning is generally applicable, we have\\ $\magnitudeI : \magnitudeII :: M' \magnitudeI : M' \magnitudeII$. @@ -9562,13 +9562,13 @@ \chapter*{Axioms} \begin{center} Let $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$, then $\magnitudeI : \magnitudeIII :: \magnitudeII : \magnitudeIV$. -For $M \magnitudeI : M \magnitudeII :: \magnitudeI : \magnitudeII$ \bycref{prop:V.XV},\\ -and $M \magnitudeI : M \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{\hypref} and \bycref{prop:V.XI};\\ -also $m \magnitudeIII : m \magnitudeIV :: \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XV}; +For $M \magnitudeI : M \magnitudeII :: \magnitudeI : \magnitudeII$ \byref{prop:V.XV},\\ +and $M \magnitudeI : M \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{\hypref} and \byref{prop:V.XI};\\ +also $m \magnitudeIII : m \magnitudeIV :: \magnitudeIII : \magnitudeIV$ \byref{prop:V.XV}; -$\therefore M \magnitudeI : M \magnitudeII :: m \magnitudeIII : m \magnitudeIV$ \bycref{prop:V.XIV},\\ +$\therefore M \magnitudeI : M \magnitudeII :: m \magnitudeIII : m \magnitudeIV$ \byref{prop:V.XIV},\\ and $\therefore$ if $M \magnitudeI >, = \mbox { or } < m \magnitudeIII$,\\ -then will $M \magnitudeII >, =, \mbox{ or } < m \magnitudeIV$ \bycref{prop:V.XIV}; +then will $M \magnitudeII >, =, \mbox{ or } < m \magnitudeIV$ \byref{prop:V.XIV}; therefore, by the fifth definition,\\ $\magnitudeI : \magnitudeIII :: \magnitudeII : \magnitudeIV$ @@ -9617,15 +9617,15 @@ \chapter*{Axioms} Take $\magnitudeI > m \magnitudeII$ to each add $M \magnitudeII$,\\ then we have $M \magnitudeI + M \magnitudeII > m \magnitudeII + M \magnitudeII$,\\ or $M (\magnitudeI + \magnitudeII) > (m + M) \magnitudeII$:\\ -but $\because \magnitudeI + \magnitudeII: \magnitudeII :: \magnitudeIII + \magnitudeIV: \magnitudeIV$ \bycref{\hypref},\\ %because +but $\because \magnitudeI + \magnitudeII: \magnitudeII :: \magnitudeIII + \magnitudeIV: \magnitudeIV$ \byref{\hypref},\\ %because and $M (\magnitudeI + \magnitudeII) > (m + M) \magnitudeII$;\\ -$\therefore M(\magnitudeIII + \magnitudeIV) > (m + M) \magnitudeIV$ \bycref{def:V.V};\\ +$\therefore M(\magnitudeIII + \magnitudeIV) > (m + M) \magnitudeIV$ \byref{def:V.V};\\ $\therefore M \magnitudeIII + M \magnitudeIV > m \magnitudeIV + M \magnitudeIV$;\\ $\therefore M \magnitudeIII > m \magnitudeIV$, by taking $M \magnitudeIV$ from both sides:\\ that is, when $M \magnitudeI > m \magnitudeII$, then $M \magnitudeIII > m \magnitudeIV$. In the same manner it may be proved, that if $M \magnitudeI = \mbox{ or } < m \magnitudeII$, then will $M \magnitudeIII = \mbox{ or } < m \magnitudeIV$;\\ -and $\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{def:V.V} +and $\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{def:V.V} $\therefore$ If magnitudes taken jointly, \&c. \end{center} @@ -9674,12 +9674,12 @@ \chapter*{Axioms} for if not, let $\magnitudeI + \magnitudeII : \magnitudeII :: \magnitudeIII + \drawMagnitude{V} : \magnitudeV$,\\ supposing $\magnitudeV \neq \magnitudeIV$; %\mbox{ not } = -$\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeV$ \bycref{prop:V.XVII}\\ -but $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{\hypref}; +$\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeV$ \byref{prop:V.XVII}\\ +but $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{\hypref}; -$\therefore \magnitudeIII : \magnitudeV :: \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XI}; +$\therefore \magnitudeIII : \magnitudeV :: \magnitudeIII : \magnitudeIV$ \byref{prop:V.XI}; -$\therefore \magnitudeV = \magnitudeIV$ \bycref{prop:V.IX},\\ +$\therefore \magnitudeV = \magnitudeIV$ \byref{prop:V.IX},\\ which is contrary to the supposition; $\therefore \magnitudeV \mbox{ is not unequal to } \magnitudeIV$;\\ @@ -9711,9 +9711,9 @@ \chapter*{Axioms} again $\magnitudeII : \magnitudeIV :: \magnitudeI : \magnitudeIII$ (alter.), -but $\magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV :: \magnitudeI : \magnitudeIII$ \bycref{\hypref}; +but $\magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV :: \magnitudeI : \magnitudeIII$ \byref{\hypref}; -therefore $\magnitudeII : \magnitudeIV :: \magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV$ \bycref{prop:V.XI}. +therefore $\magnitudeII : \magnitudeIV :: \magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV$ \byref{prop:V.XI}. $\therefore$ If a whole magnitude be to a whole, \&c. \end{center} @@ -9766,7 +9766,7 @@ \chapter*{Axioms} \def\varO{\textcolor{byyellow}{O}} \def\varP{\textcolor{byblue}{P}} \def\varQ{\textcolor{byblue}{Q}} -\enquote{Ex \ae quali,} from equality. This term is used simply by itself, when the first magnitude is to the second of the first rank, as the first to the second of the other rank; and as the second to the third of the first rank, so is the second to the third of the other; and so in order: and the inference is as mentioned in the preceding definition; whence this is called ordinate proposition. It is demonstrated in \bycref{prop:V.XXII}. +\enquote{Ex \ae quali,} from equality. This term is used simply by itself, when the first magnitude is to the second of the first rank, as the first to the second of the other rank; and as the second to the third of the first rank, so is the second to the third of the other; and so in order: and the inference is as mentioned in the preceding definition; whence this is called ordinate proposition. It is demonstrated in \byref{prop:V.XXII}. \begin{center} Thus, if there be two ranks of magnitudes,\\ @@ -9789,7 +9789,7 @@ \chapter*{Axioms} \def\varO{\textcolor{byblue}{O}} \def\varP{\textcolor{byred}{P}} \def\varQ{\textcolor{byred}{Q}} -\enquote{Ex \ae quali in proportione perturbatâ seu inordinatâ,} from equality in perturbate, or disorderly proportion. This term is used when the first magnitude is to the second of the first rank as the last but one is to the last of the second rank; and as the second is to the third of the first rank, so is the last but two to the last but one of the second rank; and as the third is to the fourth of the first rank, so is the third from the last to the last but two of the second rank; and so on in a cross order: and the inference is in the 18th definition. It is demonstrated in \bycref{prop:V.XXIII}. +\enquote{Ex \ae quali in proportione perturbatâ seu inordinatâ,} from equality in perturbate, or disorderly proportion. This term is used when the first magnitude is to the second of the first rank as the last but one is to the last of the second rank; and as the second is to the third of the first rank, so is the last but two to the last but one of the second rank; and as the third is to the fourth of the first rank, so is the third from the last to the last but two of the second rank; and so on in a cross order: and the inference is in the 18th definition. It is demonstrated in \byref{prop:V.XXIII}. \begin{center} Thus, if there be two ranks of magnitudes,\\ @@ -9824,10 +9824,10 @@ \chapter*{Axioms} $\magnitudeI : \magnitudeIV :: \magnitudeII : \magnitudeV$,\\ and $\magnitudeII : \magnitudeV :: \magnitudeIII : \magnitudeVI$ -$\therefore \magnitudeI : \magnitudeIV :: \magnitudeIII : \magnitudeVI$ \bycref{prop:V.XI}; +$\therefore \magnitudeI : \magnitudeIV :: \magnitudeIII : \magnitudeVI$ \byref{prop:V.XI}; $\therefore$ if $\magnitudeI >, =, \mbox{ or } < \magnitudeIII$,\\ -then will $\magnitudeIV >, =, \mbox{ or } < \magnitudeVI$ \bycref{prop:V.XIV}. +then will $\magnitudeIV >, =, \mbox{ or } < \magnitudeVI$ \byref{prop:V.XIV}. $\therefore$ If there be three magnitudes, \&c. \end{center} @@ -9857,23 +9857,23 @@ \chapter*{Axioms} First, let $\magnitudeI \mbox{ be } > \magnitudeIII$:\\ then, because \magnitudeII\ is any other magnitude,\\ -$\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \bycref{prop:V.VIII};\\ -but $\magnitudeV : \magnitudeVI :: \magnitudeI : \magnitudeII$ \bycref{\hypref};\\ -$\therefore \magnitudeV : \magnitudeVI > \magnitudeIII : \magnitudeII$ \bycref{prop:V.XIII};\\ -and $\because \magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \bycref{\hypref};\\ %because +$\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \byref{prop:V.VIII};\\ +but $\magnitudeV : \magnitudeVI :: \magnitudeI : \magnitudeII$ \byref{\hypref};\\ +$\therefore \magnitudeV : \magnitudeVI > \magnitudeIII : \magnitudeII$ \byref{prop:V.XIII};\\ +and $\because \magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \byref{\hypref};\\ %because $\therefore \magnitudeIII : \magnitudeII :: \magnitudeV : \magnitudeIV$ (inv.),\\ and it was shown that $\magnitudeV : \magnitudeVI > \magnitudeIII : \magnitudeII$, -$\therefore \magnitudeV : \magnitudeVI > \magnitudeV : \magnitudeIV$ \bycref{prop:V.XIII};\\ +$\therefore \magnitudeV : \magnitudeVI > \magnitudeV : \magnitudeIV$ \byref{prop:V.XIII};\\ $\therefore \magnitudeVI < \magnitudeIV$,\\ that is $\magnitudeIV > \magnitudeVI$. Secondly, let $\magnitudeI = \magnitudeIII$; then shall $\magnitudeIV = \magnitudeVI$.\\ For $\because \magnitudeI = \magnitudeIII$,\\ %because -$\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \bycref{prop:V.VII};\\ -but $\magnitudeI : \magnitudeII = \magnitudeV : \magnitudeVI$ \bycref{\hypref},\\ +$\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \byref{prop:V.VII};\\ +but $\magnitudeI : \magnitudeII = \magnitudeV : \magnitudeVI$ \byref{\hypref},\\ and $\magnitudeIII : \magnitudeII = \magnitudeV : \magnitudeIV$ (\hypstr\ and inv.),\\ -$\therefore \magnitudeV : \magnitudeVI = \magnitudeV : \magnitudeIV$ \bycref{prop:V.XI},\\ -$\therefore \magnitudeIV = \magnitudeVI$ \bycref{prop:V.IX}. +$\therefore \magnitudeV : \magnitudeVI = \magnitudeV : \magnitudeIV$ \byref{prop:V.XI},\\ +$\therefore \magnitudeIV = \magnitudeVI$ \byref{prop:V.IX}. Next, let $\magnitudeI \mbox{ be } < \magnitudeIII$, then $\magnitudeIV \mbox{ shall be } < \magnitudeVI$;\\ for $\magnitudeIII > \magnitudeI$,\\ @@ -9923,7 +9923,7 @@ \chapter*{Axioms} and\\ $M \magnitudeI, m \magnitudeII, N \magnitudeIII, M \magnitudeIV, m \magnitudeV, N \magnitudeVI,$\\ $\because \magnitudeI : \magnitudeII :: \magnitudeIV : \magnitudeV$,\\ %because -$\therefore M \magnitudeI : m \magnitudeII :: M \magnitudeIV : m \magnitudeV$ \bycref{prop:V.IV}. +$\therefore M \magnitudeI : m \magnitudeII :: M \magnitudeIV : m \magnitudeV$ \byref{prop:V.IV}. For the same reason\\ $m \magnitudeII : N \magnitudeIII :: m \magnitudeV : N \magnitudeVI$;\\ @@ -9933,8 +9933,8 @@ \chapter*{Axioms} which, taken two and two, have the same ratio; $\therefore$ if $M \magnitudeI >, = \mbox{ or } < N \magnitudeIII$\\ -then will $M \magnitudeIV >, = \mbox{ or } < N \magnitudeVI$, by \bycref{prop:V.XX},\\ -and $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeVI$ \bycref{def:V.V}. +then will $M \magnitudeIV >, = \mbox{ or } < N \magnitudeVI$, by \byref{prop:V.XX},\\ +and $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeVI$ \byref{def:V.V}. Next, let there be four magnitudes, \drawMagnitude{Ia}, \drawMagnitude{IIa}, \drawMagnitude{IIIa}, \drawMagnitude{IVa},\\ and other four, \drawMagnitude{Va}, \drawMagnitude{VIa}, \drawMagnitude{VIIa}, \drawMagnitude{VIIIa},\\ @@ -9985,20 +9985,20 @@ \chapter*{Axioms} \begin{center} $\magnitudeI, \magnitudeII, \magnitudeIII, \magnitudeIV, \magnitudeV, \magnitudeVI,$\\ $M \magnitudeI, M \magnitudeII, m \magnitudeIII, M \magnitudeIV, m \magnitudeV, m \magnitudeVI,$\\ -then $\magnitudeI : \magnitudeII :: M \magnitudeI : M \magnitudeII$ \bycref{prop:V.XV};\\ +then $\magnitudeI : \magnitudeII :: M \magnitudeI : M \magnitudeII$ \byref{prop:V.XV};\\ and for the same reason\\ $\magnitudeV : \magnitudeVI :: m \magnitudeV : m \magnitudeVI$;\\ -but $\magnitudeI : \magnitudeII :: \magnitudeV : \magnitudeVI$ \bycref{\hypref},\\ -$\therefore M \magnitudeI : M \magnitudeII :: \magnitudeV : \magnitudeVI$ \bycref{prop:V.XI};\\ -and $\because \magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \bycref{\hypref},\\ %because -$\therefore M \magnitudeII : m \magnitudeIII :: M \magnitudeIV : m \magnitudeV$ \bycref{prop:V.IV};\\ +but $\magnitudeI : \magnitudeII :: \magnitudeV : \magnitudeVI$ \byref{\hypref},\\ +$\therefore M \magnitudeI : M \magnitudeII :: \magnitudeV : \magnitudeVI$ \byref{prop:V.XI};\\ +and $\because \magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \byref{\hypref},\\ %because +$\therefore M \magnitudeII : m \magnitudeIII :: M \magnitudeIV : m \magnitudeV$ \byref{prop:V.IV};\\ then, because there are three magnitudes,\\ $M \magnitudeI, M \magnitudeII, m \magnitudeII$,\\ and other three $M \magnitudeIV, m \magnitudeV, m \magnitudeVI$,\\ which, taken two and two in a cross order, have the same ratio;\\ therefore, if $M \magnitudeI >, =, \mbox{ or } < m \magnitudeIII$,\\ -then will $M \magnitudeIV >, =, \mbox{ or } < m \magnitudeVI$ \bycref{prop:V.XXI},\\ -and $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeIV$ \bycref{def:V.V}. +then will $M \magnitudeIV >, =, \mbox{ or } < m \magnitudeVI$ \byref{prop:V.XXI},\\ +and $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeIV$ \byref{def:V.V}. Next, let there be four magnitudes,\\ \magnitudeI, \magnitudeII, \magnitudeIII, \magnitudeIV,\\ @@ -10054,16 +10054,16 @@ \chapter*{Axioms} and $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$,\\ then $\magnitudeI + \magnitudeV : \magnitudeII :: \magnitudeIII + \magnitudeVI : \magnitudeIV$. -For $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \bycref{\hypref},\\ -and $\magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$ \bycref{\hypref} and (invert.), +For $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \byref{\hypref},\\ +and $\magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$ \byref{\hypref} and (invert.), -$\therefore \magnitudeV : \magnitudeI :: \magnitudeVI : \magnitudeIII$ \bycref{prop:V.XXII};\\ +$\therefore \magnitudeV : \magnitudeI :: \magnitudeVI : \magnitudeIII$ \byref{prop:V.XXII};\\ and, because these magnitudes are proportionals, they are proportionals when taken jointly, -$\therefore \magnitudeI + \magnitudeV : \magnitudeV :: \magnitudeVI + \magnitudeIII : \magnitudeVI$ \bycref{prop:V.XVIII},\\ -but $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \bycref{\hypref}, +$\therefore \magnitudeI + \magnitudeV : \magnitudeV :: \magnitudeVI + \magnitudeIII : \magnitudeVI$ \byref{prop:V.XVIII},\\ +but $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \byref{\hypref}, -$\therefore \magnitudeI + \magnitudeV : \magnitudeII :: \magnitudeVI + \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XXII} +$\therefore \magnitudeI + \magnitudeV : \magnitudeII :: \magnitudeVI + \magnitudeIII : \magnitudeIV$ \byref{prop:V.XXII} $\therefore$ If the first, \&c. \end{center} @@ -10082,14 +10082,14 @@ \chapter*{Axioms} \begin{center} Let four magnitudes, $\drawMagnitude{Ia} + \drawMagnitude{III}, \drawMagnitude{IIa} + \drawMagnitude{IV}, \magnitudeIII, \mbox{ and } \magnitudeIV$, of the same kind, be proportionals, that is to say,\\ $\magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV :: \magnitudeIII : \magnitudeIV$,\\ -and let $\magnitudeIa + \magnitudeIII$ be the greatest of the four, and consequently by \bycref{prop:V.A,prop:V.XIV}, \magnitudeIV\ is the least;\\ +and let $\magnitudeIa + \magnitudeIII$ be the greatest of the four, and consequently by \byref{prop:V.A,prop:V.XIV}, \magnitudeIV\ is the least;\\ then will $\magnitudeIa + \magnitudeIII + \magnitudeIV \mbox{ be } > \magnitudeIIa + \magnitudeIV + \magnitudeIII$;\\ $\because \magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV :: \magnitudeIII : \magnitudeIV$, %because -$\therefore \magnitudeIa : \magnitudeIIa :: \magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV$ \bycref{prop:V.XIX},\\ -but $\magnitudeIa + \magnitudeIII > \magnitudeIIa + \magnitudeIV$ \bycref{\hypref}, +$\therefore \magnitudeIa : \magnitudeIIa :: \magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV$ \byref{prop:V.XIX},\\ +but $\magnitudeIa + \magnitudeIII > \magnitudeIIa + \magnitudeIV$ \byref{\hypref}, -$\therefore \magnitudeIa > \magnitudeIIa$ \bycref{prop:V.A};\\ +$\therefore \magnitudeIa > \magnitudeIIa$ \byref{prop:V.A};\\ to each of these add $\magnitudeIII + \magnitudeIV$,\\ $\therefore \magnitudeIa + \magnitudeIII + \magnitudeIV > \magnitudeIIa + \magnitudeIII + \magnitudeIV$. @@ -10543,7 +10543,7 @@ \chapter*{Axioms} = \dfrac {h \times k \times l \times m \times n} -{k \times l \times m \times n \times p}$ \bycref{\hypref}, +{k \times l \times m \times n \times p}$ \byref{\hypref}, or $\dfrac{\vara \times \varb}{\varb \times \varc} \times @@ -10583,7 +10583,7 @@ \chapter*{Axioms} {\vard \times \vare \times \varf \times \varg} = \dfrac {h \times k \times l \times m \times n} -{k \times l \times m \times n \times p}$ \bycref{\hypref}, +{k \times l \times m \times n \times p}$ \byref{\hypref}, and $\dfrac {m \times n} @@ -10591,7 +10591,7 @@ \chapter*{Axioms} = \dfrac {e \times f} -{f \times g}$ \bycref{\hypref}, +{f \times g}$ \byref{\hypref}, $\therefore \dfrac {h \times k \times l \times m \times n} @@ -10746,7 +10746,7 @@ \chapter*{Definitions} draw byNamedLine(MB,BC); draw byLabelsOnPolygon(A, C, M)(ALL_LABELS, 0); } -thus formed are all equal to one another, since their bases are equal. \bycref{prop:I.XXXVIII} +thus formed are all equal to one another, since their bases are equal. \byref{prop:I.XXXVIII} $\therefore$ \trianglesAMC\ and its base are respectively equimultiples of \polygonABC\ and the base \drawUnitLine{BC}. @@ -10761,10 +10761,10 @@ \chapter*{Definitions} $\therefore$ if $m$ or $6$ times \polygonABC\ $>, = \mbox{ or } < n$ or $5$ times \polygonACD\ then $m$ or $6$ times \drawUnitLine{BC} $>, = \mbox{ or } < n$ or $5$ times \drawUnitLine{CD}, $m$ and $n$ stand for every multiple taken as in the fifth definition of the Fifth Book. Although we have only shown that this property exists when $m$ equal $6$, and $n$ equal $5$, yet it is evident that the property holds good for every multiple value that may be given to $m$, and to $n$. \begin{center} -$\therefore \polygonABC\ : \polygonACD\ :: \drawUnitLine{BC} : \drawUnitLine{CD}$ \bycref{def:V.V} +$\therefore \polygonABC\ : \polygonACD\ :: \drawUnitLine{BC} : \drawUnitLine{CD}$ \byref{def:V.V} \end{center} -Parallelograms having the same altitude are the doubles of the triangles, on their bases, and are proportional to them (Part I), and hence their doubles, the parallelograms, are as their bases \bycref{prop:V.XV}. +Parallelograms having the same altitude are the doubles of the triangles, on their bases, and are proportional to them (Part I), and hence their doubles, the parallelograms, are as their bases \byref{prop:V.XV}. \qed @@ -10845,12 +10845,12 @@ \chapter*{Definitions} and $\drawLine[middle][triangleBDE]{DE,BE,DB} = -\drawLine[middle][triangleCDE]{EC,CD,DE}$ \bycref{prop:I.XXXVII}; +\drawLine[middle][triangleCDE]{EC,CD,DE}$ \byref{prop:I.XXXVII}; $\therefore \triangleBDE\ : -\drawLine[middle][triangleADE]{AD,AE,DE} :: \triangleCDE\ : \triangleADE$ \bycref{prop:V.VII};\\ -but $\triangleBDE\ : \triangleADE\ :: \drawUnitLine{DB} : \drawUnitLine{AD}$ \bycref{prop:VI.I},\\ -$\therefore \drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \bycref{prop:V.XI}. +\drawLine[middle][triangleADE]{AD,AE,DE} :: \triangleCDE\ : \triangleADE$ \byref{prop:V.VII};\\ +but $\triangleBDE\ : \triangleADE\ :: \drawUnitLine{DB} : \drawUnitLine{AD}$ \byref{prop:VI.I},\\ +$\therefore \drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \byref{prop:V.XI}. \end{center} \vfill\pagebreak @@ -10866,13 +10866,13 @@ \chapter*{Definitions} \because \drawUnitLine{DB} : \drawUnitLine{AD} &:: \triangleBDE\ : \triangleADE \\ %\mbox{ because } \mbox{ and } \drawUnitLine{EC} : \drawUnitLine{AE} &:: \triangleCDE\ : \triangleADE \\ \end{aligned} -\right\}\mbox{ \bycref{prop:VI.I} }$;\\ -but $\drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \bycref{\hypref}\\ -$\therefore \triangleBDE\ : \triangleADE :: \triangleCDE\ : \triangleADE$ \bycref{prop:V.XI}\\ -$\therefore \triangleBDE\ = \triangleCDE$ \bycref{prop:V.IX}; +\right\}\mbox{ \byref{prop:VI.I} }$;\\ +but $\drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \byref{\hypref}\\ +$\therefore \triangleBDE\ : \triangleADE :: \triangleCDE\ : \triangleADE$ \byref{prop:V.XI}\\ +$\therefore \triangleBDE\ = \triangleCDE$ \byref{prop:V.IX}; but they are on the same base \drawUnitLine{BC}, and at the same side of it, and\\ -$\therefore \drawUnitLine{DE} \parallel \drawUnitLine{BC}$ \bycref{prop:I.XXXIX}. +$\therefore \drawUnitLine{DE} \parallel \drawUnitLine{BC}$ \byref{prop:I.XXXIX}. \end{center} \qed @@ -10910,15 +10910,15 @@ \chapter*{Definitions} \startsubproposition{Part I.} \begin{center} Draw $\drawUnitLine{CE} \parallel \drawUnitLine{AD}$, to meet \drawUnitLine{AE};\\ -then, $\drawAngle{BAD} = \drawAngle{E}$ \bycref{prop:I.XXIX}. +then, $\drawAngle{BAD} = \drawAngle{E}$ \byref{prop:I.XXIX}. $\therefore \drawAngle{DAC} = \drawAngle{E}$; but $\drawAngle{DAC} = \drawAngle{C}$, $\therefore \drawAngle{C} = \drawAngle{E}$,\\ -$\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \bycref{prop:I.VI};\\ +$\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \byref{prop:I.VI};\\ and $\because \drawUnitLine{AD} \parallel \drawUnitLine{CE}$,\\ %because -$\drawUnitLine{AE} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \bycref{prop:VI.II}; +$\drawUnitLine{AE} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \byref{prop:VI.II}; but $\drawUnitLine{AE} = \drawUnitLine{CA}$;\\ -$\therefore \drawUnitLine{CA} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \bycref{prop:V.VII}. +$\therefore \drawUnitLine{CA} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \byref{prop:V.VII}. \end{center} \vfill\pagebreak @@ -10926,15 +10926,15 @@ \chapter*{Definitions} \startsubproposition{Part II.} \begin{center} Let the same construction remain,\\ -and $\drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{BD} : \drawUnitLine{DC}$ \bycref{prop:VI.II}; +and $\drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{BD} : \drawUnitLine{DC}$ \byref{prop:VI.II}; -but $\drawUnitLine{BD} : \drawUnitLine{DC} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \bycref{\hypref}\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \bycref{prop:V.XI}. +but $\drawUnitLine{BD} : \drawUnitLine{DC} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \byref{\hypref}\\ +$\therefore \drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \byref{prop:V.XI}. -and $\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \bycref{prop:V.IX},\\ -and $\therefore \drawAngle{E} = \drawAngle{C}$ \bycref{prop:I.V};\\ +and $\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \byref{prop:V.IX},\\ +and $\therefore \drawAngle{E} = \drawAngle{C}$ \byref{prop:I.V};\\ but since $\drawUnitLine{AD} \parallel \drawUnitLine{CE}$; $\drawAngle{DAC} = \drawAngle{C}$,\\ -and $\drawAngle{BAD} = \drawAngle{E}$ \bycref{prop:I.XXIX}; +and $\drawAngle{BAD} = \drawAngle{E}$ \byref{prop:I.XXIX}; $\therefore \drawAngle{C} = \drawAngle{E}$, and $\drawAngle{BAD} = \drawAngle{DAC}$,\\ and $\therefore$ \drawUnitLine{AD} bisects \drawAngle{BAD,DAC}. @@ -10982,7 +10982,7 @@ \chapter*{Definitions} \begin{center} Draw \drawUnitLine{AF} and \drawUnitLine{DF}. -Then, $\because \drawAngle{BCA} = \drawAngle{E}$, $\drawUnitLine{CA} \parallel \drawUnitLine{DF,DE}$ \bycref{prop:I.XXVIII}; %because +Then, $\because \drawAngle{BCA} = \drawAngle{E}$, $\drawUnitLine{CA} \parallel \drawUnitLine{DF,DE}$ \byref{prop:I.XXVIII}; %because and for a like reason $\drawUnitLine{CD} \parallel \drawUnitLine{AB,AF}$, @@ -10996,12 +10996,12 @@ \chapter*{Definitions} } is a parallelogram. -But $\drawUnitLine{BC} : \drawUnitLine{EC} :: \drawUnitLine{DF} : \drawUnitLine{DE}$ \bycref{prop:VI.II}; +But $\drawUnitLine{BC} : \drawUnitLine{EC} :: \drawUnitLine{DF} : \drawUnitLine{DE}$ \byref{prop:VI.II}; -and since $\drawUnitLine{DF} = \drawUnitLine{CA}$ \bycref{prop:I.XXXIV},\\ +and since $\drawUnitLine{DF} = \drawUnitLine{CA}$ \byref{prop:I.XXXIV},\\ $\drawUnitLine{BC} : \drawUnitLine{EC} :: \drawUnitLine{CA} : \drawUnitLine{DE}$; -and by alternation $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EC} : \drawUnitLine{DE}$ \bycref{prop:V.XVI}. +and by alternation $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EC} : \drawUnitLine{DE}$ \byref{prop:V.XVI}. In like manner it may be shown, that\\ $\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{BC} : \drawUnitLine{EC}$;\\ @@ -11010,7 +11010,7 @@ \chapter*{Definitions} but it has been already proved that\\ $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EC} : \drawUnitLine{DE}$\\ and therefore, ex \ae quali,\\ -$\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{CD} : \drawUnitLine{DE}$ \bycref{prop:V.XXII},\\ +$\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{CD} : \drawUnitLine{DE}$ \byref{prop:V.XXII},\\ therefore the sides about the equal angles are proportional, and those which are opposite to the equal angles are homologous. \end{center} @@ -11060,21 +11060,21 @@ \chapter*{Definitions} } \begin{center} From the extremities of \drawUnitLine{EF}, draw \drawUnitLine{FG} and \drawUnitLine{GE}, making\\ -$\drawAngle{GEF} = \drawAngle{B}$, $\drawAngle{EFG} = \drawAngle{C}$ \bycref{prop:I.XXIII};\\ -and consequently $\drawAngle{G} = \drawAngle{A}$ \bycref{prop:I.XXXII},\\ +$\drawAngle{GEF} = \drawAngle{B}$, $\drawAngle{EFG} = \drawAngle{C}$ \byref{prop:I.XXIII};\\ +and consequently $\drawAngle{G} = \drawAngle{A}$ \byref{prop:I.XXXII},\\ and since the triangles are equiangular,\\ -$\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{GE} : \drawUnitLine{EF}$ \bycref{prop:VI.IV};\\ -but $\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{DE} : \drawUnitLine{EF}$ \bycref{\hypref};\\ +$\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{GE} : \drawUnitLine{EF}$ \byref{prop:VI.IV};\\ +but $\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{DE} : \drawUnitLine{EF}$ \byref{\hypref};\\ $\therefore \drawUnitLine{DE} : \drawUnitLine{EF} :: \drawUnitLine{GE} : \drawUnitLine{EF}$, -and consequently $\drawUnitLine{DE} = \drawUnitLine{GE}$ \bycref{prop:V.IX}. +and consequently $\drawUnitLine{DE} = \drawUnitLine{GE}$ \byref{prop:V.IX}. In like manner it may be shown that\\ $\drawUnitLine{FD} = \drawUnitLine{FG}$. Therefore, the two triangles having a common base \drawUnitLine{EF}, and their sides equal, have also equal angles opposite to equal sides, i.\ e.\\ -$\drawAngle{DEF} = \drawAngle{GEF}$ and $\drawAngle{EFD} = \drawAngle{EFG}$ \bycref{prop:I.VIII}.\\ -but $\drawAngle{GEF} = \drawAngle{B}$ \bycref{\constref} and $\therefore \drawAngle{DEF} = \drawAngle{B}$;\\ +$\drawAngle{DEF} = \drawAngle{GEF}$ and $\drawAngle{EFD} = \drawAngle{EFG}$ \byref{prop:I.VIII}.\\ +but $\drawAngle{GEF} = \drawAngle{B}$ \byref{\constref} and $\therefore \drawAngle{DEF} = \drawAngle{B}$;\\ for the same reason $\drawAngle{EFD} = \drawAngle{C}$,\\ -and consequently $\drawAngle{D} = \drawAngle{A}$ \bycref{prop:I.XXXII};\\ +and consequently $\drawAngle{D} = \drawAngle{A}$ \byref{prop:I.XXXII};\\ and therefore the triangles are equiangular, and it is evident that the homologous sides subtended by the equal angles. \end{center} @@ -11123,18 +11123,18 @@ \chapter*{Definitions} draw \drawUnitLine{GD} and \drawUnitLine{FG},\\ making $\drawAngle{GFD} = \drawAngle{C}$,\\ and $\drawAngle{FDG} = \drawAngle{A}$;\\ -then $\drawAngle{G} = \drawAngle{B}$ \bycref{prop:I.XXXII},\\ +then $\drawAngle{G} = \drawAngle{B}$ \byref{prop:I.XXXII},\\ and two triangles being equiangular,\\ -$\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{FG} : \drawUnitLine{FD}$ \bycref{prop:VI.IV};\\ -but $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \bycref{\hypref}\\ -$\therefore \drawUnitLine{FG} : \drawUnitLine{FD} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \bycref{prop:V.XI},\\ -and consequently $\drawUnitLine{FG} = \drawUnitLine{EF}$ \bycref{prop:V.IX};\\ +$\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{FG} : \drawUnitLine{FD}$ \byref{prop:VI.IV};\\ +but $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \byref{\hypref}\\ +$\therefore \drawUnitLine{FG} : \drawUnitLine{FD} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \byref{prop:V.XI},\\ +and consequently $\drawUnitLine{FG} = \drawUnitLine{EF}$ \byref{prop:V.IX};\\ $\therefore \triangleDEF\ = -\drawLine[bottom][triangleDGF]{FD,FG,GD}$ in every respect \bycref{prop:I.IV}.\\ -But $\drawAngle{GFD} = \drawAngle{A}$ \bycref{\constref},\\ +\drawLine[bottom][triangleDGF]{FD,FG,GD}$ in every respect \byref{prop:I.IV}.\\ +But $\drawAngle{GFD} = \drawAngle{A}$ \byref{\constref},\\ and $\therefore \drawAngle{EFD} = \drawAngle{A}$;\\ and since also $\drawAngle{EFD} = \drawAngle{C}$,\\ -$\drawAngle{E} = \drawAngle{B}$ \bycref{prop:I.XXXII};\\ +$\drawAngle{E} = \drawAngle{B}$ \byref{prop:I.XXXII};\\ and $\therefore$ \triangleABC\ and \triangleDEF\ are equiangular, with their equal angles opposite to homologous sides. \end{center} @@ -11180,25 +11180,25 @@ \chapter*{Definitions} \begin{center} First let it be assumed that the angles \drawAngle{A} and \drawAngle{D} are each less than a right angle: then if it be supposed that \drawAngle{ABG,GBC} and \drawAngle{E} contained by the proportional sides, are not equal, let \drawAngle{ABG,GBC} be greater, and make $\drawAngle{GBC} = \drawAngle{E}$. -$\because \drawAngle{C} = \drawAngle{F}$ \bycref{\hypref} and $\drawAngle{GBC} = \drawAngle{E}$ \bycref{\constref}\\ %Because -$\therefore \drawAngle{CGB} = \drawAngle{D}$ \bycref{prop:I.XXXII}; +$\because \drawAngle{C} = \drawAngle{F}$ \byref{\hypref} and $\drawAngle{GBC} = \drawAngle{E}$ \byref{\constref}\\ %Because +$\therefore \drawAngle{CGB} = \drawAngle{D}$ \byref{prop:I.XXXII}; -$\therefore \drawUnitLine{BC} : \drawUnitLine{BG} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \bycref{prop:VI.IV},\\ -but $\drawUnitLine{BC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \bycref{\hypref} +$\therefore \drawUnitLine{BC} : \drawUnitLine{BG} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \byref{prop:VI.IV},\\ +but $\drawUnitLine{BC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \byref{\hypref} $\therefore \drawUnitLine{BC} : \drawUnitLine{BG} :: \drawUnitLine{BC} : \drawUnitLine{AB}$;\\ -$\therefore \drawUnitLine{BG} = \drawUnitLine{AB}$ \bycref{prop:V.IX},\\ -and $\therefore \drawAngle{A} = \drawAngle{BGA}$ \bycref{prop:I.V}. +$\therefore \drawUnitLine{BG} = \drawUnitLine{AB}$ \byref{prop:V.IX},\\ +and $\therefore \drawAngle{A} = \drawAngle{BGA}$ \byref{prop:I.V}. -But \drawAngle{A} is less than a right angle \bycref{\hypref}\\ +But \drawAngle{A} is less than a right angle \byref{\hypref}\\ $\therefore$ \drawAngle{BGA} is less than a right angle;\\ -and $\therefore$ \drawAngle{CGB} must be greater than a right angle \bycref{prop:I.XIII}, but it has been proven $= \drawAngle{D}$ and therefore less than a right angle, which is absurd. $\therefore$ \drawAngle{ABG,GBC} and \drawAngle{E} are not unequal; +and $\therefore$ \drawAngle{CGB} must be greater than a right angle \byref{prop:I.XIII}, but it has been proven $= \drawAngle{D}$ and therefore less than a right angle, which is absurd. $\therefore$ \drawAngle{ABG,GBC} and \drawAngle{E} are not unequal; -$\therefore$ they are equal, and since $\drawAngle{C} = \drawAngle{F}$ \bycref{\hypref}\\ -$\therefore \drawAngle{A} = \drawAngle{D}$ \bycref{prop:I.XXXII}, and therefore the triangles are equiangular. +$\therefore$ they are equal, and since $\drawAngle{C} = \drawAngle{F}$ \byref{\hypref}\\ +$\therefore \drawAngle{A} = \drawAngle{D}$ \byref{prop:I.XXXII}, and therefore the triangles are equiangular. \end{center} -But if \drawAngle{A} and \drawAngle{D} be assumed to be each not less than a right angle, it may be proved as before, that the triangles are equiangular, and have the sides about equal the angles proportional \bycref{prop:VI.IV}. +But if \drawAngle{A} and \drawAngle{D} be assumed to be each not less than a right angle, it may be proved as before, that the triangles are equiangular, and have the sides about equal the angles proportional \byref{prop:VI.IV}. \qed @@ -11225,11 +11225,11 @@ \chapter*{Definitions} \problem{I}{n}{a right angled tiangle (\drawPolygon[bottom][triangleABC]{ABD,ADC}), if a perpendicular (\drawUnitLine{AD}) be drawn from the right angle to the opposite side, the triangles (\drawPolygon[bottom][triangleABD]{ABD}, \drawPolygon[bottom][triangleADC]{ADC}) on each side of it are similar to the whole triangle and to each other.} \begin{center} -$\because \drawAngle{DAB,CAD} = \drawAngle{D}$ \bycref{ax:I.XI},\\ +$\because \drawAngle{DAB,CAD} = \drawAngle{D}$ \byref{ax:I.XI},\\ and \drawAngle{B} common to \triangleABC\ and \triangleABD;\\ % Because -$\drawAngle{C} = \drawAngle{DAB}$ \bycref{prop:I.XXXII}; +$\drawAngle{C} = \drawAngle{DAB}$ \byref{prop:I.XXXII}; -$\therefore$ \triangleABC\ and \triangleABD\ are equiangular; and consequently have their sides about the equal angles proportional \bycref{prop:VI.IV}, and are therefore similar \bycref{def:VI.I}. +$\therefore$ \triangleABC\ and \triangleABD\ are equiangular; and consequently have their sides about the equal angles proportional \byref{prop:VI.IV}, and are therefore similar \byref{def:VI.I}. In like manner it may be proved that \triangleADC\ is similar to \triangleABC; but \triangleABD\ has been shown to be similar to \triangleABC;\\ $\therefore$ \triangleABD\ and \triangleADC\ are similar to the whole and to each other. @@ -11275,11 +11275,11 @@ \chapter*{Definitions} \drawSizedLine{AF} is the required part of \drawSizedLine{AF,FB}. For since $\drawSizedLine{DF} \parallel \drawSizedLine{BE}$\\ -$\drawSizedLine{AF} : \drawSizedLine{FB} :: \drawSizedLine{AD} : \drawSizedLine{DC,CE}$ \bycref{prop:VI.II},\\ -and by composition \bycref{prop:V.XVIII};\\ +$\drawSizedLine{AF} : \drawSizedLine{FB} :: \drawSizedLine{AD} : \drawSizedLine{DC,CE}$ \byref{prop:VI.II},\\ +and by composition \byref{prop:V.XVIII};\\ $\drawSizedLine{AF,FB} : \drawSizedLine{AF} :: \drawSizedLine{AD,DC,CE} : \drawSizedLine{AD}$; -but \drawSizedLine{AD,DC,CE} contains \drawSizedLine{AD} as often as \drawSizedLine{AF,FB} contains the required part \bycref{\constref}; +but \drawSizedLine{AD,DC,CE} contains \drawSizedLine{AD} as often as \drawSizedLine{AF,FB} contains the required part \byref{\constref}; $\therefore$ \drawSizedLine{AF} is the required part. \end{center} @@ -11329,7 +11329,7 @@ \chapter*{Definitions} \begin{center} From either extremity of the given line \drawSizedLine{AF,FG,GB} draw \drawSizedLine{AD,DE,EC} making any angle; -take \drawSizedLine{AD}, \drawSizedLine{DE} and \drawSizedLine{EC} equal to \drawSizedLine{A'D'}, \drawSizedLine{D'E'} and \drawSizedLine{E'C'} respectively \bycref{prop:I.II}; +take \drawSizedLine{AD}, \drawSizedLine{DE} and \drawSizedLine{EC} equal to \drawSizedLine{A'D'}, \drawSizedLine{D'E'} and \drawSizedLine{E'C'} respectively \byref{prop:I.II}; draw \drawSizedLine{CB}, and draw \drawSizedLine{EG} and \drawSizedLine{DF} $\parallel$ to it. @@ -11339,10 +11339,10 @@ \chapter*{Definitions} \nointerlineskip\hbox{\drawSizedLine{EG}} \nointerlineskip\hbox{\drawSizedLine{DF}}}\right\}$ are $\parallel$,\\ -$\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{EC} : \drawSizedLine{DE}$ \bycref{prop:VI.II},\\ -or $\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{E'C'} : \drawSizedLine{D'E'}$ \bycref{\constref},\\ -and $\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{DE} : \drawSizedLine{AD}$ \bycref{prop:VI.II},\\ -$\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{D'E'} : \drawSizedLine{A'D'}$ \bycref{\constref}, +$\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{EC} : \drawSizedLine{DE}$ \byref{prop:VI.II},\\ +or $\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{E'C'} : \drawSizedLine{D'E'}$ \byref{\constref},\\ +and $\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{DE} : \drawSizedLine{AD}$ \byref{prop:VI.II},\\ +$\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{D'E'} : \drawSizedLine{A'D'}$ \byref{\constref}, and $\therefore$ the given line \drawSizedLine{AF,FG,GB} is divided similarly to \drawSizedLine{A'D',D'E',E'C'}. \end{center} @@ -11383,13 +11383,13 @@ \chapter*{Definitions} At either extremity of the given line \drawSizedLine{AB} draw \drawSizedLine{AC,CE} making an angle;\\ take $\drawSizedLine{AC} = \drawSizedLine{A'C'}$, and draw \drawSizedLine{BC};\\ make $\drawSizedLine{BD} = \drawSizedLine{A'C'}$,\\ -and draw $\drawSizedLine{DE} \parallel \drawSizedLine{BC}$; \bycref{prop:I.XXXI}\\ +and draw $\drawSizedLine{DE} \parallel \drawSizedLine{BC}$; \byref{prop:I.XXXI}\\ \drawSizedLine{CE} is the third proportional to \drawSizedLine{AB} and \drawSizedLine{A'C'}. For since $\drawSizedLine{BC} \parallel \drawSizedLine{DE}$,\\ -$\therefore \drawSizedLine{AB} : \drawSizedLine{BD} :: \drawSizedLine{AC} : \drawSizedLine{CE}$ \bycref{prop:VI.II};\\ -but $\drawSizedLine{BD} = \drawSizedLine{AC} = \drawSizedLine{A'C'}$ \bycref{\constref};\\ -$\therefore \drawSizedLine{AB} : \drawSizedLine{A'C'} :: \drawSizedLine{A'C'} : \drawSizedLine{CE}$ \bycref{prop:V.VII}. +$\therefore \drawSizedLine{AB} : \drawSizedLine{BD} :: \drawSizedLine{AC} : \drawSizedLine{CE}$ \byref{prop:VI.II};\\ +but $\drawSizedLine{BD} = \drawSizedLine{AC} = \drawSizedLine{A'C'}$ \byref{\constref};\\ +$\therefore \drawSizedLine{AB} : \drawSizedLine{A'C'} :: \drawSizedLine{A'C'} : \drawSizedLine{CE}$ \byref{prop:V.VII}. \end{center} \qed @@ -11438,11 +11438,11 @@ \chapter*{Definitions} and $\drawUnitLine{GE} = \drawUnitLine{B}$,\\ also $\drawUnitLine{DH} = \drawUnitLine{C}$,\\ draw \drawUnitLine{GH},\\ -and $\drawUnitLine{EF} \parallel \drawUnitLine{GH}$ \bycref{prop:I.XXXI};\\ +and $\drawUnitLine{EF} \parallel \drawUnitLine{GH}$ \byref{prop:I.XXXI};\\ \drawSizedLine{HF} is the fourth proportional. On account of the parallels,\\ -$\drawUnitLine{DG} : \drawUnitLine{GE} :: \drawUnitLine{DH} : \drawUnitLine{HF}$ \bycref{prop:VI.II};\\ +$\drawUnitLine{DG} : \drawUnitLine{GE} :: \drawUnitLine{DH} : \drawUnitLine{HF}$ \byref{prop:VI.II};\\ but $\left\{\vcenter{ \nointerlineskip\hbox{\drawSizedLine{A}} @@ -11453,9 +11453,9 @@ \chapter*{Definitions} \nointerlineskip\hbox{\drawSizedLine{DG}} \nointerlineskip\hbox{\drawSizedLine{GE}} \nointerlineskip\hbox{\drawSizedLine{DH}}}\right\}$ -\bycref{\constref}; +\byref{\constref}; -$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{C} : \drawUnitLine{HF}$ \bycref{prop:V.VII}. +$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{C} : \drawUnitLine{HF}$ \byref{prop:V.VII}. \end{center} \qed @@ -11515,10 +11515,10 @@ \chapter*{Definitions} Draw \drawUnitLine{AD} and \drawUnitLine{CD}. -Since \drawAngle{D} is a right angle \bycref{prop:III.XXXI},\\ +Since \drawAngle{D} is a right angle \byref{prop:III.XXXI},\\ and \drawUnitLine{BD} is $\perp$ from it upon the opposite side,\\ -$\therefore$ \drawUnitLine{BD} is a mean proportional between \drawUnitLine{AB} and \drawUnitLine{BC} \bycref{prop:VI.VIII},\\ -and $\therefore$ between \drawUnitLine{A} and \drawUnitLine{B} \bycref{\constref}. +$\therefore$ \drawUnitLine{BD} is a mean proportional between \drawUnitLine{AB} and \drawUnitLine{BC} \byref{prop:VI.VIII},\\ +and $\therefore$ between \drawUnitLine{A} and \drawUnitLine{B} \byref{\constref}. \end{center} \qed @@ -11554,26 +11554,26 @@ \chapter*{Definitions} And parallelograms which have one angle in each equal, and the sides about them reciprocally proportional, are equal.} \begin{center} -Let \drawUnitLine{BG} and \drawUnitLine{BF}; and \drawUnitLine{BD} and \drawUnitLine{BE}, be so placed that \drawUnitLine{BG,BF} and \drawUnitLine{BD,BE} may be continued right lines. It is evident that they may assume this position. \bycref{prop:I.XIII,prop:I.XIV,prop:I.XV} +Let \drawUnitLine{BG} and \drawUnitLine{BF}; and \drawUnitLine{BD} and \drawUnitLine{BE}, be so placed that \drawUnitLine{BG,BF} and \drawUnitLine{BD,BE} may be continued right lines. It is evident that they may assume this position. \byref{prop:I.XIII,prop:I.XIV,prop:I.XV} Complete \drawPolygon{EHFB}. Since $\drawPolygon{CEBG} = \drawPolygon{BFAD}$; -$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \bycref{prop:V.VII}\\ -$\therefore \drawUnitLine{BG} : \drawUnitLine{BF} :: \drawUnitLine{BD} : \drawUnitLine{BE}$ \bycref{prop:VI.I} +$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \byref{prop:V.VII}\\ +$\therefore \drawUnitLine{BG} : \drawUnitLine{BF} :: \drawUnitLine{BD} : \drawUnitLine{BE}$ \byref{prop:VI.I} The same construction remaining:\\ $\drawUnitLine{BG} : \drawUnitLine{BF} :: \left\{ \begin{aligned} - \drawPolygon{CEBG} &: \drawPolygon{EHFB} \mbox{\bycref{prop:VI.I}}\\ - \drawUnitLine{BD} &: \drawUnitLine{BE} \mbox{\bycref{\hypref}}\\ - \drawPolygon{BFAD} &: \drawPolygon{EHFB} \mbox{\bycref{prop:VI.I}}\\ + \drawPolygon{CEBG} &: \drawPolygon{EHFB} \mbox{\byref{prop:VI.I}}\\ + \drawUnitLine{BD} &: \drawUnitLine{BE} \mbox{\byref{\hypref}}\\ + \drawPolygon{BFAD} &: \drawPolygon{EHFB} \mbox{\byref{prop:VI.I}}\\ \end{aligned} \right.$\\ -$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \bycref{prop:V.XI}\\ -and $\therefore \drawPolygon{CEBG} = \drawPolygon{BFAD}$ \bycref{prop:V.IX}. +$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \byref{prop:V.XI}\\ +and $\therefore \drawPolygon{CEBG} = \drawPolygon{BFAD}$ \byref{prop:V.IX}. \end{center} \qed @@ -11606,18 +11606,18 @@ \chapter*{Definitions} And two triangles which have an angle of the one equal to an angle of the other, and the sides about the equal angles reciprocally proportional, are equal.} \startsubproposition{I.} -Let the triangles be so placed that the equal angles \drawAngle{DAE} and \drawAngle{BAC} may be vertically opposite, that is to say, so that \drawUnitLine{AE} and \drawUnitLine{BA} may be in the same straight line. Whence also \drawUnitLine{AC} and \drawUnitLine{DA} must be in the same straight line \bycref{prop:I.XIV}. +Let the triangles be so placed that the equal angles \drawAngle{DAE} and \drawAngle{BAC} may be vertically opposite, that is to say, so that \drawUnitLine{AE} and \drawUnitLine{BA} may be in the same straight line. Whence also \drawUnitLine{AC} and \drawUnitLine{DA} must be in the same straight line \byref{prop:I.XIV}. \begin{center} Draw \drawUnitLine{BD}, then\\ $\begin{aligned} \drawUnitLine{DA} : \drawUnitLine{BA} -&:: \drawPolygon{DAE} : \drawPolygon{DAB} \mbox{\bycref{prop:VI.I}}\\ -&:: \drawPolygon{ABC} : \drawPolygon{DAB} \mbox{\bycref{prop:V.VII}}\\ -&:: \drawUnitLine{AC} : \drawUnitLine{DA} \mbox{\bycref{prop:VI.I}}\\ +&:: \drawPolygon{DAE} : \drawPolygon{DAB} \mbox{\byref{prop:VI.I}}\\ +&:: \drawPolygon{ABC} : \drawPolygon{DAB} \mbox{\byref{prop:V.VII}}\\ +&:: \drawUnitLine{AC} : \drawUnitLine{DA} \mbox{\byref{prop:VI.I}}\\ \end{aligned}$ -$\therefore \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$ \bycref{prop:V.XI}. +$\therefore \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$ \byref{prop:V.XI}. \end{center} \vfill\pagebreak @@ -11625,14 +11625,14 @@ \chapter*{Definitions} \startsubproposition{II.} \begin{center} Let the same construction remain, and\\ -$\drawPolygon{DAE} : \drawPolygon{DAB} :: \drawUnitLine{DA} : \drawUnitLine{BA}$ \bycref{prop:VI.I}\\ -and $\drawUnitLine{AC} : \drawUnitLine{DA} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \bycref{prop:VI.I} +$\drawPolygon{DAE} : \drawPolygon{DAB} :: \drawUnitLine{DA} : \drawUnitLine{BA}$ \byref{prop:VI.I}\\ +and $\drawUnitLine{AC} : \drawUnitLine{DA} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \byref{prop:VI.I} -but $ \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$, \bycref{\hypref} +but $ \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$, \byref{\hypref} -$\therefore \drawPolygon{DAE} : \drawPolygon{DAB} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \bycref{prop:V.XI}; +$\therefore \drawPolygon{DAE} : \drawPolygon{DAB} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \byref{prop:V.XI}; -$\therefore \drawPolygon{DAE} = \drawPolygon{ABC}$ \bycref{prop:V.IX}. +$\therefore \drawPolygon{DAE} = \drawPolygon{ABC}$ \byref{prop:V.IX}. \end{center} \qed @@ -11682,9 +11682,9 @@ \chapter*{Definitions} complete the parallelograms \drawPolygon{ABFG} and \drawPolygon{CDEH}. and since,\\ -$\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \bycref{\hypref}\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \bycref{\constref}\\ -$\therefore \drawPolygon{ABFG} = \drawPolygon{CDEH}$ \bycref{prop:VI.XIV},\\ +$\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \byref{\hypref}\\ +$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \byref{\constref}\\ +$\therefore \drawPolygon{ABFG} = \drawPolygon{CDEH}$ \byref{prop:VI.XIV},\\ that is, the rectangle contained be the extremes, equal to the rectangle contained be the means. \end{center} @@ -11695,11 +11695,11 @@ \chapter*{Definitions} Let the same construction remain;\\ $\because \drawUnitLine{F} = \drawUnitLine{AG}$, $\drawPolygon{ABFG} = \drawPolygon{CDEH}$\\ %because and $\drawUnitLine{CH} = \drawUnitLine{E}$.\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \bycref{prop:VI.XIV} +$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \byref{prop:VI.XIV} But $\drawUnitLine{CH} = \drawUnitLine{E}$,\\ -and $\drawUnitLine{AG} = \drawUnitLine{F}$ \bycref{\constref}\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \bycref{prop:V.VII}. +and $\drawUnitLine{AG} = \drawUnitLine{F}$ \byref{\constref}\\ +$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \byref{prop:V.VII}. \end{center} \qed @@ -11751,7 +11751,7 @@ \chapter*{Definitions} \mbox{ then } \drawUnitLine{A} : \drawUnitLine{B} &:: \drawUnitLine{D} : \drawUnitLine{C}, \\ \therefore \drawUnitLine{A} \times \drawUnitLine{C} &= \drawUnitLine{B} \times \drawUnitLine{D} \\ \end{aligned}$\\ -\bycref{prop:VI.XVI}. +\byref{prop:VI.XVI}. But $\drawUnitLine{D} = \drawUnitLine{B}$,\\ $\therefore \drawUnitLine{B} \times \drawUnitLine{D} = \drawUnitLine{B} \times \drawUnitLine{B} \mbox{ or } = \drawUnitLine{B}^2$;\\ @@ -11762,7 +11762,7 @@ \chapter*{Definitions} \begin{center} Assume $\drawUnitLine{D} = \drawUnitLine{B}$,\\ then $\drawUnitLine{A} \times \drawUnitLine{C} = \drawUnitLine{D} \times \drawUnitLine{B}$,\\ -$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{D} : \drawUnitLine{C}$ \bycref{prop:VI.XVI},\\ +$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{D} : \drawUnitLine{C}$ \byref{prop:VI.XVI},\\ and $\drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{B} : \drawUnitLine{C}$. \end{center} @@ -11834,19 +11834,19 @@ \chapter*{Definitions} Then $\polygonCD = \drawPolygon[middle][polygonAB]{BAG,BGH,BHK}$. -It is evident from the construction and \bycref{prop:I.XXXII} that the figures are equiangular; and since the triangles \drawPolygon{DCF} and \drawPolygon{BAG} are equiangular;\\ -then by \bycref{prop:VI.IV}, +It is evident from the construction and \byref{prop:I.XXXII} that the figures are equiangular; and since the triangles \drawPolygon{DCF} and \drawPolygon{BAG} are equiangular;\\ +then by \byref{prop:VI.IV}, $\drawUnitLine{AB}:\drawUnitLine{AG} :: \drawUnitLine{CD} : \drawUnitLine{CF}$\\ and $\drawUnitLine{AG}:\drawUnitLine{BG} :: \drawUnitLine{CF} : \drawUnitLine{DF}$ Again, because \drawPolygon{DFE} and \drawPolygon{BGH} are equiangular,\\ $\drawUnitLine{BG}:\drawUnitLine{GH} :: \drawUnitLine{DF} : \drawUnitLine{FE}$\\ $\therefore$ ex \ae quali,\\ -$\drawUnitLine{AG}:\drawUnitLine{GH} :: \drawUnitLine{CF} : \drawUnitLine{FE}$ \bycref{prop:V.XXII} % improvement: Byrne had VI.XXII wrongly referenced here +$\drawUnitLine{AG}:\drawUnitLine{GH} :: \drawUnitLine{CF} : \drawUnitLine{FE}$ \byref{prop:V.XXII} % improvement: Byrne had VI.XXII wrongly referenced here In like manner it may be shown that the remaining sides of the two figures are proportional. -$\therefore$ by \bycref{prop:VI.I}\\ +$\therefore$ by \byref{prop:VI.I}\\ \polygonAB\ is similar to \polygonCD\ and similarly situated; and on the given line \drawUnitLine{AB}. \end{center} @@ -11890,17 +11890,17 @@ \chapter*{Definitions} \begin{center} $\drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{EF} : \drawUnitLine{BG}$;\\ draw \drawUnitLine{AG}.\\ -$\drawUnitLine{BG,GC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \bycref{prop:VI.IV};\\ -$\therefore \drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{AB} : \drawUnitLine{DE}$ \bycref{prop:V.XVI},\\ -but $\drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{EF} : \drawUnitLine{BG}$ \bycref{\constref},\\ +$\drawUnitLine{BG,GC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \byref{prop:VI.IV};\\ +$\therefore \drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{AB} : \drawUnitLine{DE}$ \byref{prop:V.XVI},\\ +but $\drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{EF} : \drawUnitLine{BG}$ \byref{\constref},\\ $\therefore \drawUnitLine{EF} : \drawUnitLine{BG} :: \drawUnitLine{AB} : \drawUnitLine{DE}$\\ -consequently $\triangleDEF = \drawPolygon[bottom][triangleABG]{ABG}$ for they have the sides about the equal angles \drawAngle{E} and \drawAngle{B} reciprocally proportional \bycref{prop:VI.XV}; +consequently $\triangleDEF = \drawPolygon[bottom][triangleABG]{ABG}$ for they have the sides about the equal angles \drawAngle{E} and \drawAngle{B} reciprocally proportional \byref{prop:VI.XV}; -$\therefore \triangleABC : \triangleDEF :: \triangleABC : \triangleABG$ \bycref{prop:V.VII};\\ -but $\triangleABC : \triangleABG :: \drawUnitLine{BG,GC} : \drawUnitLine{BG}$ \bycref{prop:VI.I}, +$\therefore \triangleABC : \triangleDEF :: \triangleABC : \triangleABG$ \byref{prop:V.VII};\\ +but $\triangleABC : \triangleABG :: \drawUnitLine{BG,GC} : \drawUnitLine{BG}$ \byref{prop:VI.I}, $\therefore \triangleABC : \triangleDEF :: \drawUnitLine{BG,GC} : \drawUnitLine{BG}$,\\ -that is to say, the triangles are to one another in the duplicate ratio of their homologous sides \drawUnitLine{EF} and \drawUnitLine{BG,GC} \bycref{def:V.XI}. +that is to say, the triangles are to one another in the duplicate ratio of their homologous sides \drawUnitLine{EF} and \drawUnitLine{BG,GC} \byref{def:V.XI}. \end{center} \qed @@ -11958,7 +11958,7 @@ \chapter*{Definitions} \begin{center} $\therefore$ \drawPolygon{BCD} and \drawPolygon{GHK} are similar,\\ -and $\drawAngle{BDC} = \drawAngle{GKH}$ \bycref{prop:VI.VI}; +and $\drawAngle{BDC} = \drawAngle{GKH}$ \byref{prop:VI.VI}; but $\drawAngle{BDE,BDC} = \drawAngle{GKL,GKH}$ because they are angles of similar polygons; therefore the remainders \drawAngle{BDE} and \drawAngle{GKL} are equal;\\ hence $\drawUnitLine{BD} : \drawUnitLine{CD} :: \drawUnitLine{GK} : \drawUnitLine{HK}$,\\ @@ -11966,20 +11966,20 @@ \chapter*{Definitions} and $\drawUnitLine{CD} : \drawUnitLine{DE} :: \drawUnitLine{HK} : \drawUnitLine{KL}$,\\ on account of the similar polygons,\\ $\therefore \drawUnitLine{BD} : \drawUnitLine{DE} :: \drawUnitLine{GK} : \drawUnitLine{KL}$,\\ -ex \ae quali \bycref{prop:V.XXII}, and as these proportional sides contain equal angles, the triangles \drawPolygon{BDE} and \drawPolygon{GKL} are similar \bycref{prop:VI.VI}. +ex \ae quali \byref{prop:V.XXII}, and as these proportional sides contain equal angles, the triangles \drawPolygon{BDE} and \drawPolygon{GKL} are similar \byref{prop:VI.VI}. In like manner it may be shown that the triangles \drawPolygon{BEA} and \drawPolygon{GLF} are similar. -But \drawPolygon{BCD} is to \drawPolygon{GHK} in the duplicate ratio of \drawUnitLine{BD} to \drawUnitLine{GK} \bycref{prop:VI.XIX},\\ +But \drawPolygon{BCD} is to \drawPolygon{GHK} in the duplicate ratio of \drawUnitLine{BD} to \drawUnitLine{GK} \byref{prop:VI.XIX},\\ and \drawPolygon{BDE} is to \drawPolygon{GKL} in like manner, in the duplicate ratio of \drawUnitLine{BD} to \drawUnitLine{GK};\\ -$\therefore \drawPolygon{BCD} : \drawPolygon{GHK} :: \drawPolygon{BDE} : \drawPolygon{GKL}$ \bycref{prop:V.XI}. +$\therefore \drawPolygon{BCD} : \drawPolygon{GHK} :: \drawPolygon{BDE} : \drawPolygon{GKL}$ \byref{prop:V.XI}. Again \drawPolygon{BDE} is to \drawPolygon{GKL} in the duplicate ratio of \drawUnitLine{BE} to \drawUnitLine{GL}, and \drawPolygon{BEA} is to \drawPolygon{GLF} in the duplicate ratio of \drawUnitLine{BE} to \drawUnitLine{GL}.\\ $\begin{aligned} \drawPolygon{BCD} : \drawPolygon{GHK} &:: \drawPolygon{BDE} : \drawPolygon{GKL} \\ &:: \drawPolygon{BEA} : \drawPolygon{GLF} \\ \end{aligned}$;\\ -and as one of the antecedents is to one of the consequents, so is the sum of all the antecedents to the sum of all the consequents; that is to say, the similar triangles have to one another the same ratio as the polygons \bycref{prop:V.XII}. +and as one of the antecedents is to one of the consequents, so is the sum of all the antecedents to the sum of all the consequents; that is to say, the similar triangles have to one another the same ratio as the polygons \byref{prop:V.XII}. But \drawPolygon{BCD} is to \drawPolygon{GHK} in the duplicate ratio of \drawUnitLine{BC} to \drawUnitLine{GH};\\ $\therefore$ \drawPolygon{BEA,BDE,BCD} is to \drawPolygon{GLF,GKL,GHK} in the duplicate ratio of \drawUnitLine{BC} to \drawUnitLine{GH}. @@ -12012,7 +12012,7 @@ \chapter*{Definitions} \drawCurrentPictureInMargin \problem{R}{ectilinear}{figures (\drawPolygon[bottom]{A} and \drawPolygon[bottom]{B}) which are similar to the same figure (\drawPolygon[bottom]{C}) are similar also to each other.} -Since \polygonA\ and \polygonC\ are similar, they are equiangular, and have the sides about the equal angles proportional \bycref{def:VI.I}; and since the figures \polygonB\ and \polygonC\ are also similar, they are equiangular, and have the sides about the equal angles proportional; therefore \polygonA\ and \polygonB\ are also equiangular, and have the sides about the equal angles proportional \bycref{prop:V.XI}, and are therefore similar. +Since \polygonA\ and \polygonC\ are similar, they are equiangular, and have the sides about the equal angles proportional \byref{def:VI.I}; and since the figures \polygonB\ and \polygonC\ are also similar, they are equiangular, and have the sides about the equal angles proportional; therefore \polygonA\ and \polygonB\ are also equiangular, and have the sides about the equal angles proportional \byref{prop:V.XI}, and are therefore similar. \qed @@ -12064,18 +12064,18 @@ \chapter*{Definitions} \startsubproposition{Part I.} \begin{center} -Take \drawUnitLine{O} a third proportional to \drawUnitLine{AB} and \drawUnitLine{CD}, and \drawUnitLine{P} a third proportional to \drawUnitLine{EF} and \drawUnitLine{GH} \bycref{prop:VI.XI}; +Take \drawUnitLine{O} a third proportional to \drawUnitLine{AB} and \drawUnitLine{CD}, and \drawUnitLine{P} a third proportional to \drawUnitLine{EF} and \drawUnitLine{GH} \byref{prop:VI.XI}; -since $\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \bycref{\hypref}\\ -$\drawUnitLine{CD} : \drawUnitLine{O} :: \drawUnitLine{GH} : \drawUnitLine{P}$ \bycref{\constref} +since $\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \byref{\hypref}\\ +$\drawUnitLine{CD} : \drawUnitLine{O} :: \drawUnitLine{GH} : \drawUnitLine{P}$ \byref{\constref} $\therefore$ ex \ae quali,\\ $\drawUnitLine{AB} : \drawUnitLine{O} :: \drawUnitLine{EF} : \drawUnitLine{P}$; -but $\drawPolygon{AB} : \drawPolygon{CD} :: \drawUnitLine{AB} : \drawUnitLine{O}$ \bycref{prop:VI.XX},\\ +but $\drawPolygon{AB} : \drawPolygon{CD} :: \drawUnitLine{AB} : \drawUnitLine{O}$ \byref{prop:VI.XX},\\ and $\drawPolygon{EF} : \drawPolygon{GH} :: \drawUnitLine{EF} : \drawUnitLine{P}$; -$\therefore \drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \bycref{prop:V.XI}. +$\therefore \drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \byref{prop:V.XI}. \end{center} \vfill\pagebreak @@ -12084,11 +12084,11 @@ \chapter*{Definitions} \begin{center} Let the same construction remain; -$\drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \bycref{\hypref}, +$\drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \byref{\hypref}, -$\therefore \drawUnitLine{AB} : \drawUnitLine{O} :: \drawUnitLine{EF} : \drawUnitLine{P}$ \bycref{\constref}, +$\therefore \drawUnitLine{AB} : \drawUnitLine{O} :: \drawUnitLine{EF} : \drawUnitLine{P}$ \byref{\constref}, -and $\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \bycref{prop:V.XI}. +and $\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \byref{prop:V.XI}. \end{center} \qed @@ -12128,14 +12128,14 @@ \chapter*{Definitions} \begin{center} Since $\drawAngle{BCD} + \drawAngle{DCG} = \drawTwoRightAngles$,\\ -and $\drawAngle{GCE} = \drawAngle{BCD}$ \bycref{\hypref},\\ +and $\drawAngle{GCE} = \drawAngle{BCD}$ \byref{\hypref},\\ $\drawAngle{GCE} + \drawAngle{DCG} = \drawTwoRightAngles$,\\ -and $\therefore$ \drawUnitLine{CE} and \drawUnitLine{DC} form one straight line \bycref{prop:I.XIV}; +and $\therefore$ \drawUnitLine{CE} and \drawUnitLine{DC} form one straight line \byref{prop:I.XIV}; complete \drawPolygon[bottom]{CDHG}. -Since $\polygonABCD\ : \polygonCDHG\ :: \drawUnitLine{BC} : \drawUnitLine{CG}$ \bycref{prop:VI.I},\\ -and $\polygonCDHG\ : \polygonEFGC\ :: \drawUnitLine{DC} : \drawUnitLine{CE}$ \bycref{prop:VI.I},\\ +Since $\polygonABCD\ : \polygonCDHG\ :: \drawUnitLine{BC} : \drawUnitLine{CG}$ \byref{prop:VI.I},\\ +and $\polygonCDHG\ : \polygonEFGC\ :: \drawUnitLine{DC} : \drawUnitLine{CE}$ \byref{prop:VI.I},\\ \polygonABCD\ has to \polygonEFGC\ a ratio compounded of the ratios of \drawUnitLine{BC} to \drawUnitLine{CG}, and of \drawUnitLine{DC} to \drawUnitLine{CE}. \end{center} @@ -12214,7 +12214,7 @@ \chapter*{Definitions} stopAutoLabeling; stopGlobalRotation; } -are similar \bycref{prop:VI.IV},\\ +are similar \byref{prop:VI.IV},\\ $\therefore \drawUnitLine{GA} : \drawUnitLine{GF} :: \drawUnitLine{GA,DG} : \drawUnitLine{DK,KC}$; and the remaining opposite sides are equal to those,\\ @@ -12293,23 +12293,23 @@ \chapter*{Definitions} stopAutoLabeling; stopTempAngleScale; } = \polygonD$,\\ -and having $\drawAngle{B} = \drawAngle{C}$ \bycref{prop:I.XLV},\\ -and then \drawUnitLine{BC} and \drawUnitLine{CF} will lie in the same straight line \bycref{prop:I.XXIX,prop:I.XIV}. +and having $\drawAngle{B} = \drawAngle{C}$ \byref{prop:I.XLV},\\ +and then \drawUnitLine{BC} and \drawUnitLine{CF} will lie in the same straight line \byref{prop:I.XXIX,prop:I.XIV}. -Between \drawUnitLine{BC} and \drawUnitLine{CF} find a mean proportional \drawUnitLine{GH} \bycref{prop:VI.XIII},\\ +Between \drawUnitLine{BC} and \drawUnitLine{CF} find a mean proportional \drawUnitLine{GH} \byref{prop:VI.XIII},\\ and upon \drawUnitLine{GH} describe \drawPolygon[bottom][triangleKGH]{KGH}, similar to \triangleABC, and similarly situated. Then $\triangleKGH\ = \polygonD$. For since \triangleABC\ and \triangleKGH\ are similar, and\\ -$\drawUnitLine{BC} : \drawUnitLine{GH} :: \drawUnitLine{GH} : \drawUnitLine{CF}$ \bycref{\constref},\\ -$\triangleABC\ : \triangleKGH\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \bycref{prop:VI.XX}; - -but $\polygonBCEL\ : \polygonCFME\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \bycref{prop:VI.I};\\ -$\therefore \triangleABC\ : \triangleKGH\ :: \polygonBCEL\ : \polygonCFME$ \bycref{prop:V.XI};\\ -but $\triangleABC\ = \polygonBCEL$ \bycref{\constref},\\ -and $\therefore \triangleKGH\ = \polygonCFME$ \bycref{prop:V.XIV};\\ -and $\polygonCFME\ = \polygonD$ \bycref{\constref};\\ +$\drawUnitLine{BC} : \drawUnitLine{GH} :: \drawUnitLine{GH} : \drawUnitLine{CF}$ \byref{\constref},\\ +$\triangleABC\ : \triangleKGH\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \byref{prop:VI.XX}; + +but $\polygonBCEL\ : \polygonCFME\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \byref{prop:VI.I};\\ +$\therefore \triangleABC\ : \triangleKGH\ :: \polygonBCEL\ : \polygonCFME$ \byref{prop:V.XI};\\ +but $\triangleABC\ = \polygonBCEL$ \byref{\constref},\\ +and $\therefore \triangleKGH\ = \polygonCFME$ \byref{prop:V.XIV};\\ +and $\polygonCFME\ = \polygonD$ \byref{\constref};\\ consequently, \triangleKGH\ which is similar to \triangleABC\ is also $= \polygonD$. \end{center} @@ -12379,14 +12379,14 @@ \chapter*{Definitions} draw byLabelsOnPolygon(A, H, C, noPoint)(ALL_LABELS, 0); stopGlobalRotation; } -be the diagonal of \parallelogramABCD\ and draw $\drawUnitLine{KH} \parallel \drawUnitLine{AG}$ \bycref{prop:I.XXXI}. +be the diagonal of \parallelogramABCD\ and draw $\drawUnitLine{KH} \parallel \drawUnitLine{AG}$ \byref{prop:I.XXXI}. -Since \drawLine[bottom][parallelogramAKHG]{AG,GH,KH,AK} and \parallelogramABCD\ are about the same diagonal \lineAHC, and have \drawAngle{A} common, they are similar \bycref{prop:VI.XXIV}; +Since \drawLine[bottom][parallelogramAKHG]{AG,GH,KH,AK} and \parallelogramABCD\ are about the same diagonal \lineAHC, and have \drawAngle{A} common, they are similar \byref{prop:VI.XXIV}; $\therefore \drawSizedLine{AG} : \drawSizedLine{AK} :: \drawSizedLine{AG,GD} : \drawSizedLine{AK,KE,EB}$;\\ -but $\drawSizedLine{AG} : \drawSizedLine{AK,KE} :: \drawSizedLine{AG,GD} : \drawSizedLine{AK,KE,EB}$ \bycref{\hypref},\\ +but $\drawSizedLine{AG} : \drawSizedLine{AK,KE} :: \drawSizedLine{AG,GD} : \drawSizedLine{AK,KE,EB}$ \byref{\hypref},\\ $\therefore \drawSizedLine{AG} : \drawSizedLine{AK} :: \drawSizedLine{AG} : \drawSizedLine{AK,KE}$,\\ -and $\therefore \drawSizedLine{AK} = \drawSizedLine{AK,KE}$ \bycref{prop:V.IX}, which is absurd. +and $\therefore \drawSizedLine{AK} = \drawSizedLine{AK,KE}$ \byref{prop:V.IX}, which is absurd. $\therefore$ \lineAHC\ is not the diagonal of \parallelogramABCD\ in the same manner it can be demonstrated that no other line is except \drawUnitLine{AC}. \end{center} @@ -12428,7 +12428,7 @@ \chapter*{Definitions} then $\drawPolygon[bottom]{ADGKEF,CDGK} > \drawPolygon[bottom]{BCKH,CDGK}$. \end{center} -For it has been demonstrated already \bycref{prop:II.V}, that the square of half the line is equal to the rectangle contained by any unequal segments together with the square of the part intermediate between the middle point and the point of unequal section. The square described on half the line exceeds therefore the rectangle contained by any unequal segments of the line. +For it has been demonstrated already \byref{prop:II.V}, that the square of half the line is equal to the rectangle contained by any unequal segments together with the square of the part intermediate between the middle point and the point of unequal section. The square described on half the line exceeds therefore the rectangle contained by any unequal segments of the line. \qed @@ -12475,21 +12475,21 @@ \chapter*{Definitions} problem is solved. But if $\drawSizedLine{AE,EC}^2 \neq \drawSizedLine{G}^2$,\\ -then must $\drawSizedLine{AE,EC} > \drawSizedLine{G}$ \bycref{\hypref}. +then must $\drawSizedLine{AE,EC} > \drawSizedLine{G}$ \byref{\hypref}. Draw $\drawSizedLine{DC} \perp \drawSizedLine{AE,EC} = \drawSizedLine{G}$;\\ make $\drawSizedLine{DC,CF} = \drawSizedLine{AE,EC} \mbox{ or } \drawSizedLine{CB}$;\\ with \drawSizedLine{DC,CF} as radius describe a circle cutting the given line; draw \drawSizedLine{DE}. -Then $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} + \drawSizedLine{EC}^2 = \drawSizedLine{AE,EC}^2$ \bycref{prop:II.V} $= \drawSizedLine{DE}^2$. +Then $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} + \drawSizedLine{EC}^2 = \drawSizedLine{AE,EC}^2$ \byref{prop:II.V} $= \drawSizedLine{DE}^2$. -But $\drawSizedLine{DE}^2 = \drawSizedLine{DC}^2 + \drawSizedLine{EC}^2$ \bycref{prop:I.XLVII}; +But $\drawSizedLine{DE}^2 = \drawSizedLine{DC}^2 + \drawSizedLine{EC}^2$ \byref{prop:I.XLVII}; $\therefore \drawSizedLine{AE} \times \drawSizedLine{EC,CB} + \drawSizedLine{EC}^2 = \drawSizedLine{DC}^2 + \drawSizedLine{EC}^2$,\\ from both, take $\drawSizedLine{EC}^2$,\\ and $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} = \drawSizedLine{DC}^2$. -But $\drawSizedLine{DC} = \drawSizedLine{G}$ \bycref{\constref},\\ +But $\drawSizedLine{DC} = \drawSizedLine{G}$ \byref{\constref},\\ and $\therefore$ \drawSizedLine{AE,EC,CB} is so divided that $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} = \drawSizedLine{G}^2$. \end{center} @@ -12529,9 +12529,9 @@ \chapter*{Definitions} draw \drawSizedLine{CD};\\ and with the radius \drawSizedLine{CD}, describe a circle meeting \drawSizedLine{AC,CB} produced. -Then $\drawSizedLine{AC,CB,BF} \times \drawSizedLine{BF} + \drawSizedLine{CB}^2 = \drawSizedLine{AC,CB}^2$ \bycref{prop:II.VI} $= \drawSizedLine{CD}^2$. +Then $\drawSizedLine{AC,CB,BF} \times \drawSizedLine{BF} + \drawSizedLine{CB}^2 = \drawSizedLine{AC,CB}^2$ \byref{prop:II.VI} $= \drawSizedLine{CD}^2$. -But $\drawSizedLine{CD}^2 = \drawSizedLine{BD}^2 + \drawSizedLine{CB}^2$ \bycref{prop:I.XLVII} +But $\drawSizedLine{CD}^2 = \drawSizedLine{BD}^2 + \drawSizedLine{CB}^2$ \byref{prop:I.XLVII} $\therefore \drawSizedLine{AC,CB,BF} \times \drawSizedLine{BF} + \drawSizedLine{CB}^2 = \drawSizedLine{BD}^2 + \drawSizedLine{CB}^2$,\\ from both take $\drawSizedLine{CB}^2$,\\ @@ -12577,11 +12577,11 @@ \chapter*{Definitions} \problem{T}{o}{cut a given finite straight line (\drawProportionalLine{AE,EB}) in extreme and mean ratio.} \begin{center} -On \drawProportionalLine{AE,EB} describe the square \drawPolygon[middle][squareABHC]{ACFE,EFHB} \bycref{prop:I.XLVI};\\ -and produce \drawProportionalLine{CA}, so that $\drawProportionalLine{CA,AG} \times \drawProportionalLine{AG} = \drawProportionalLine{AE,EB}^2$ \bycref{prop:VI.XXIX};\\ +On \drawProportionalLine{AE,EB} describe the square \drawPolygon[middle][squareABHC]{ACFE,EFHB} \byref{prop:I.XLVI};\\ +and produce \drawProportionalLine{CA}, so that $\drawProportionalLine{CA,AG} \times \drawProportionalLine{AG} = \drawProportionalLine{AE,EB}^2$ \byref{prop:VI.XXIX};\\ take $\drawProportionalLine{AE} = \drawProportionalLine{AG}$,\\ and draw $\drawProportionalLine{DE} \parallel \drawProportionalLine{CA,AG}$,\\ -meeting $\drawProportionalLine{GD} \parallel \drawProportionalLine{AE,EB}$ \bycref{prop:I.XXXI}. +meeting $\drawProportionalLine{GD} \parallel \drawProportionalLine{AE,EB}$ \byref{prop:I.XXXI}. Then $\drawFromCurrentPicture[middle][rectangleCFDG]{ draw byNamedPolygon(ACFE); @@ -12594,7 +12594,7 @@ \chapter*{Definitions} will be $= \drawPolygon{EFHB}$, which is $= \drawProportionalLine{AE,EB} \times \drawProportionalLine{EB}$;\\ that is $\drawProportionalLine{AE}^2 = \drawProportionalLine{AE,EB} \times \drawProportionalLine{EB}$;\\ $\therefore \drawProportionalLine{AE,EB} : \drawProportionalLine{AE} :: \drawProportionalLine{AE} : \drawProportionalLine{EB}$,\\ -and \drawProportionalLine{AE,EB} is divided in extreme and mean ratio \bycref{def:VI.III}. +and \drawProportionalLine{AE,EB} is divided in extreme and mean ratio \byref{def:VI.III}. \end{center} \qed @@ -12636,7 +12636,7 @@ \chapter*{Definitions} \begin{center} From the right angle draw \drawProportionalLine{AD} perpendicular to \drawProportionalLine{BD,DC};\\ -then $\drawProportionalLine{BD,DC} : \drawProportionalLine{CA} :: \drawProportionalLine{CA} : \drawProportionalLine{DC}$ \bycref{prop:VI.VIII}. +then $\drawProportionalLine{BD,DC} : \drawProportionalLine{CA} :: \drawProportionalLine{CA} : \drawProportionalLine{DC}$ \byref{prop:VI.VIII}. $\therefore \drawFromCurrentPicture[bottom][figBC]{ @@ -12652,7 +12652,7 @@ \chapter*{Definitions} draw byNamedPolygon(CA); stopAutoLabeling; stopGlobalRotation; -} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{DC}$ \bycref{prop:VI.XX}.\\ +} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{DC}$ \byref{prop:VI.XX}.\\ but $\figBC\ : \drawFromCurrentPicture[bottom][figAB]{ startGlobalRotation(-angle(A-B)); @@ -12660,7 +12660,7 @@ \chapter*{Definitions} draw byNamedPolygon(AB); stopAutoLabeling; stopGlobalRotation; -} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{BD}$ \bycref{prop:VI.XX}. +} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{BD}$ \byref{prop:VI.XX}. Hence $\drawProportionalLine{DC} + \drawProportionalLine{BD} : \drawProportionalLine{BD,DC} :: \figAB\ + \figCA\ : \figBC$;\\ but $\drawProportionalLine{DC} + \drawProportionalLine{BD} = \drawProportionalLine{BD,DC}$;\\ @@ -12722,18 +12722,18 @@ \chapter*{Definitions} \begin{center} Since $\drawUnitLine{AB} \parallel \drawUnitLine{DC}$,\\ -$\drawAngle{A} = \drawAngle{ACD}$ \bycref{prop:I.XXIX};\\ +$\drawAngle{A} = \drawAngle{ACD}$ \byref{prop:I.XXIX};\\ and also since $\drawUnitLine{CA} \parallel \drawUnitLine{ED}$,\\ -$\drawAngle{ACD} = \drawAngle{D}$ \bycref{prop:I.XXIX};\\ +$\drawAngle{ACD} = \drawAngle{D}$ \byref{prop:I.XXIX};\\ $\therefore \drawAngle{A} = \drawAngle{D}$;\\ -and since $\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{DC} : \drawUnitLine{ED}$ \bycref{\hypref},\\ -the triangles are equiangular \bycref{prop:VI.VI}; +and since $\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{DC} : \drawUnitLine{ED}$ \byref{\hypref},\\ +the triangles are equiangular \byref{prop:VI.VI}; $\therefore \drawAngle{B} = \drawAngle{DCE}$;\\ but $\drawAngle{A} = \drawAngle{ACD}$; -$\therefore \drawAngle{BCA} + \drawAngle{ACD} + \drawAngle{DCE} = \drawAngle{BCA} + \drawAngle{A} + \drawAngle{B} = \drawTwoRightAngles$ \bycref{prop:I.XXXII},\\ -and $\therefore$ \drawUnitLine{BC} and \drawUnitLine{CE} lie in the same straight line \bycref{prop:I.XIV}. +$\therefore \drawAngle{BCA} + \drawAngle{ACD} + \drawAngle{DCE} = \drawAngle{BCA} + \drawAngle{A} + \drawAngle{B} = \drawTwoRightAngles$ \byref{prop:I.XXXII},\\ +and $\therefore$ \drawUnitLine{BC} and \drawUnitLine{CE} lie in the same straight line \byref{prop:I.XIV}. \end{center} \qed @@ -12848,7 +12848,7 @@ \chapter*{Definitions} stopGlobalRotation; }, \&c.\ each $= \arcEF$, draw the radii to the extremities of the equal arcs. -The since the arcs \arcBC, \arcCK, \arcKL, \&c.\ are all equal, the angles \drawAngle{BC}, \drawAngle{CK}, \drawAngle{KL}, \&c.\ are also equal \bycref{prop:III.XXVII}; $\therefore$ \drawAngle{BC,CK,KL} is the same multiple of \drawAngle{BC} which the arc \drawFromCurrentPicture[middle][arcBL]{ +The since the arcs \arcBC, \arcCK, \arcKL, \&c.\ are all equal, the angles \drawAngle{BC}, \drawAngle{CK}, \drawAngle{KL}, \&c.\ are also equal \byref{prop:III.XXVII}; $\therefore$ \drawAngle{BC,CK,KL} is the same multiple of \drawAngle{BC} which the arc \drawFromCurrentPicture[middle][arcBL]{ startGlobalRotation(180-angle(B-L)); startAutoLabeling; draw byNamedArcSeq(0)(BC,CK,KL); @@ -12863,14 +12863,14 @@ \chapter*{Definitions} } is of the arc \arcEF. \begin{center} -Then it is evident \bycref{prop:III.XXVII},\\ +Then it is evident \byref{prop:III.XXVII},\\ if \drawAngle{BC,CK,KL} (or if $m$ times \drawAngle{BC}) $>, =, < \drawAngle{EF,FM,MN}$ (or $n$ times \drawAngle{EF})\\ then \arcBL (or $m$ times \arcBC) $>, =, < \arcEN$ (or $n$ times \arcEF); \end{center} -$\therefore \drawAngle{BC} : \drawAngle{EF} :: \arcBC\ : \arcEF$ \bycref{def:V.V}, or the angles at the centre are as the arcs on which they stand; but the angles at the circumference being halves of the angles at the centre \bycref{prop:III.XX} are in the same ratio \bycref{prop:V.XV}, and therefore are as the arcs on which they stand. +$\therefore \drawAngle{BC} : \drawAngle{EF} :: \arcBC\ : \arcEF$ \byref{def:V.V}, or the angles at the centre are as the arcs on which they stand; but the angles at the circumference being halves of the angles at the centre \byref{prop:III.XX} are in the same ratio \byref{prop:V.XV}, and therefore are as the arcs on which they stand. -It is evident, that sectors in equal circles, and on equal arcs are equal \bycref{prop:I.IV,prop:III.XXIV,prop:III.XXVII,def:III.IX}. Hence, if the sectors be substituted for the angles in the above demonstration, the second part of the proportion will be established, that is, in equal circles the sectors have the same ratio to one another as the arcs on which they stand. +It is evident, that sectors in equal circles, and on equal arcs are equal \byref{prop:I.IV,prop:III.XXIV,prop:III.XXVII,def:III.IX}. Hence, if the sectors be substituted for the angles in the above demonstration, the second part of the proportion will be established, that is, in equal circles the sectors have the same ratio to one another as the arcs on which they stand. \qed @@ -12915,14 +12915,14 @@ \chapter*{Definitions} \begin{center} For if \drawSizedLine{BE} be drawn $\parallel \drawSizedLine{DC}$,\\ $\begin{aligned} - \mbox{then } \drawAngle{DCB} &= \drawAngle{EBC} \mbox{, \bycref{prop:I.XXIX};}\\ - & = \drawAngle{FCD} \mbox{, \bycref{\hypref},}\\ - & = \drawAngle{E} \mbox{, \bycref{prop:I.XXIX};}\\ + \mbox{then } \drawAngle{DCB} &= \drawAngle{EBC} \mbox{, \byref{prop:I.XXIX};}\\ + & = \drawAngle{FCD} \mbox{, \byref{\hypref},}\\ + & = \drawAngle{E} \mbox{, \byref{prop:I.XXIX};}\\ \end{aligned}$\\ -and $\therefore \drawSizedLine{EC} = \drawSizedLine{BC}$, \bycref{prop:I.VI},\\ -and $\drawSizedLine{AE,EC} : \drawSizedLine{BC} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \bycref{prop:V.VII};\\ -But also, $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \bycref{prop:VI.II};\\ -and therefore $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{BC}$ \bycref{prop:V.XI}. +and $\therefore \drawSizedLine{EC} = \drawSizedLine{BC}$, \byref{prop:I.VI},\\ +and $\drawSizedLine{AE,EC} : \drawSizedLine{BC} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \byref{prop:V.VII};\\ +But also, $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \byref{prop:VI.II};\\ +and therefore $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{BC}$ \byref{prop:V.XI}. \end{center} \qed @@ -12963,16 +12963,16 @@ \chapter*{Definitions} Let \drawSizedLine{AD} be drawn, making $\drawAngle{CAE} = \drawAngle{EAB}$;\\ then shall $\drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{BD} \times \drawSizedLine{DC} + \drawSizedLine{AD}^2$. -About \drawLine[bottom]{CA,DC,BD,AB} describe \drawCircle[middle][1/4]{O} \bycref{prop:IV.V},\\ +About \drawLine[bottom]{CA,DC,BD,AB} describe \drawCircle[middle][1/4]{O} \byref{prop:IV.V},\\ produce \drawSizedLine{AD} to meet the circle, and draw \drawSizedLine{CE}. -Since $\drawAngle{CAE} = \drawAngle{EAB}$ \bycref{\hypref},\\ -and $\drawAngle{B} = \drawAngle{E}$ \bycref{prop:III.XXI},\\ -$\therefore$ \drawLine[bottom]{AD,BD,AB} and \drawLine[middle]{CA,CE,DE,AD} are equiangular \bycref{prop:I.XXXII};\\ -$\therefore \drawSizedLine{AB} : \drawSizedLine{AD} :: \drawSizedLine{AD,DE} : \drawSizedLine{CA}$ \bycref{prop:VI.IV};\\ -$\therefore \drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{AD} \times \drawSizedLine{AD,DE}$ \bycref{prop:VI.XVI}\\ -$= \drawSizedLine{DE} \times \drawSizedLine{AD} + \drawSizedLine{AD}^2$ \bycref{prop:II.III};\\ -but $\drawSizedLine{DE} \times \drawSizedLine{AD} = \drawSizedLine{BD} \times \drawSizedLine{DC}$ \bycref{prop:III.XXXV};\\ +Since $\drawAngle{CAE} = \drawAngle{EAB}$ \byref{\hypref},\\ +and $\drawAngle{B} = \drawAngle{E}$ \byref{prop:III.XXI},\\ +$\therefore$ \drawLine[bottom]{AD,BD,AB} and \drawLine[middle]{CA,CE,DE,AD} are equiangular \byref{prop:I.XXXII};\\ +$\therefore \drawSizedLine{AB} : \drawSizedLine{AD} :: \drawSizedLine{AD,DE} : \drawSizedLine{CA}$ \byref{prop:VI.IV};\\ +$\therefore \drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{AD} \times \drawSizedLine{AD,DE}$ \byref{prop:VI.XVI}\\ +$= \drawSizedLine{DE} \times \drawSizedLine{AD} + \drawSizedLine{AD}^2$ \byref{prop:II.III};\\ +but $\drawSizedLine{DE} \times \drawSizedLine{AD} = \drawSizedLine{BD} \times \drawSizedLine{DC}$ \byref{prop:III.XXXV};\\ $\therefore \drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{BD} \times \drawSizedLine{DC} + \drawSizedLine{AD}^2$ \end{center} @@ -13015,13 +13015,13 @@ \chapter*{Definitions} draw $\drawUnitLine{AD} \perp \drawUnitLine{BD,DC}$;\\ then shall $\drawUnitLine{AB} \times \drawUnitLine{CA} = \drawUnitLine{AD} \times $ the diameter of the described circle. -Describe \drawCircle[middle][1/4]{O} \bycref{prop:IV.V}, draw its diameter \drawUnitLine{AE}, and draw \drawUnitLine{CE};\\ -then $\because \drawAngle{D} = \drawAngle{BCA,ECB}$ \bycref{\constref,prop:III.XXXI};\\ %because -and $\drawAngle{B} = \drawAngle{E}$ \bycref{prop:III.XXI};\\ -$\therefore$ \drawLine[bottom]{AD,BD,AB} is equiangular to \drawLine[middle]{CA,CE,AE} \bycref{prop:VI.IV}; +Describe \drawCircle[middle][1/4]{O} \byref{prop:IV.V}, draw its diameter \drawUnitLine{AE}, and draw \drawUnitLine{CE};\\ +then $\because \drawAngle{D} = \drawAngle{BCA,ECB}$ \byref{\constref,prop:III.XXXI};\\ %because +and $\drawAngle{B} = \drawAngle{E}$ \byref{prop:III.XXI};\\ +$\therefore$ \drawLine[bottom]{AD,BD,AB} is equiangular to \drawLine[middle]{CA,CE,AE} \byref{prop:VI.IV}; $\therefore \drawUnitLine{AB} : \drawUnitLine{AD} :: \drawUnitLine{AE} : \drawUnitLine{CA}$;\\ -and $\therefore \drawUnitLine{AB} \times \drawUnitLine{CA} = \drawUnitLine{AD} \times \drawUnitLine{AE}$ \bycref{prop:VI.XVI}. +and $\therefore \drawUnitLine{AB} \times \drawUnitLine{CA} = \drawUnitLine{AD} \times \drawUnitLine{AE}$ \byref{prop:VI.XVI}. \end{center} \qed @@ -13069,17 +13069,17 @@ \chapter*{Definitions} and draw \drawUnitLine{BE,ED} and \drawUnitLine{AC};\\ then $\drawUnitLine{BE,ED} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD} + \drawUnitLine{DA} \times \drawUnitLine{BC}$. -Make $\drawAngle{EAB} = \drawAngle{DAC}$ \bycref{prop:I.XXIII},\\ -$\therefore \drawAngle{EAB,CAE} = \drawAngle{CAE,DAC}$; and $\drawAngle{BCA} = \drawAngle{D}$ \bycref{prop:III.XXI}; +Make $\drawAngle{EAB} = \drawAngle{DAC}$ \byref{prop:I.XXIII},\\ +$\therefore \drawAngle{EAB,CAE} = \drawAngle{CAE,DAC}$; and $\drawAngle{BCA} = \drawAngle{D}$ \byref{prop:III.XXI}; -$\therefore \drawUnitLine{DA} : \drawUnitLine{ED} :: \drawUnitLine{AC} : \drawUnitLine{BC}$ \bycref{prop:VI.IV};\\ -and $\therefore \drawUnitLine{ED} \times \drawUnitLine{AC} = \drawUnitLine{DA} \times \drawUnitLine{BC}$ \bycref{prop:VI.XVI};\\ -again, $\because \drawAngle{EAB} = \drawAngle{DAC}$ \bycref{\constref},\\ %because -and $\drawAngle{B} = \drawAngle{ACD}$ \bycref{prop:III.XXI};\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{BE} :: \drawUnitLine{AC} : \drawUnitLine{CD}$ \bycref{prop:VI.IV};\\ -and $\therefore \drawUnitLine{BE} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD}$ \bycref{prop:VI.XVI};\\ +$\therefore \drawUnitLine{DA} : \drawUnitLine{ED} :: \drawUnitLine{AC} : \drawUnitLine{BC}$ \byref{prop:VI.IV};\\ +and $\therefore \drawUnitLine{ED} \times \drawUnitLine{AC} = \drawUnitLine{DA} \times \drawUnitLine{BC}$ \byref{prop:VI.XVI};\\ +again, $\because \drawAngle{EAB} = \drawAngle{DAC}$ \byref{\constref},\\ %because +and $\drawAngle{B} = \drawAngle{ACD}$ \byref{prop:III.XXI};\\ +$\therefore \drawUnitLine{AB} : \drawUnitLine{BE} :: \drawUnitLine{AC} : \drawUnitLine{CD}$ \byref{prop:VI.IV};\\ +and $\therefore \drawUnitLine{BE} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD}$ \byref{prop:VI.XVI};\\ but, from above, $\drawUnitLine{ED} \times \drawUnitLine{AC} = \drawUnitLine{DA} \times \drawUnitLine{BC}$;\\ -$\therefore \drawUnitLine{BE,ED} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD} + \drawUnitLine{DA} \times \drawUnitLine{BC}$ \bycref{prop:II.I}. +$\therefore \drawUnitLine{BE,ED} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD} + \drawUnitLine{DA} \times \drawUnitLine{BC}$ \byref{prop:II.I}. \end{center} \qed diff --git a/byrne-ru-latex.tex b/byrne-ru-latex.tex index 4a6cd11..71d0b89 100644 --- a/byrne-ru-latex.tex +++ b/byrne-ru-latex.tex @@ -1,6 +1,5 @@ \documentclass[letters,booklanguage=russian]{byrnebook} %\usepackage{lua-visual-debug} -\usepackage[russian]{babel} \begin{document} @@ -336,7 +335,7 @@ \part*{Введение} \drawAngle{BAC} общий и~$\drawUnitLine{AB} = \drawUnitLine{AC}$. $\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$\\ -и $\drawAngle{CEB} = \drawAngle{BDC}$ \bycref{prop:I.IV}. +и $\drawAngle{CEB} = \drawAngle{BDC}$ \byref{prop:I.IV}. Теперь в~\drawFromCurrentPicture{ startAutoLabeling; @@ -353,7 +352,7 @@ \part*{Введение} и $\drawUnitLine{BE} = \drawUnitLine{CD}$. $\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ -и $\drawAngle{DCB} = \drawAngle{CBE}$ \bycref{prop:I.IV}. +и $\drawAngle{DCB} = \drawAngle{CBE}$ \byref{prop:I.IV}. Но $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, @@ -366,9 +365,9 @@ \part*{Введение} \emph{Если использовать буквы} \end{center} -Продлим равные стороны AB и~AC через концы третьей стороны BC. На любой из продленных частей BD возьмем любую точку D и~от другой отсечем AE, равную AD \bycref{prop:I.III}. Теперь взятые таким образом на продленных сторонах точки E и~D соединим прямыми линиями DC и~BE с~противолежащими концами третьей стороны треугольника. +Продлим равные стороны AB и~AC через концы третьей стороны BC. На любой из продленных частей BD возьмем любую точку D и~от другой отсечем AE, равную AD \byref{prop:I.III}. Теперь взятые таким образом на продленных сторонах точки E и~D соединим прямыми линиями DC и~BE с~противолежащими концами третьей стороны треугольника. -В треугольниках DAC и~EAB стороны DA и~AC соответственно равны EA и~AB, а~прилежащий угол A общий обоим. Следовательно \bycref{prop:I.IV}, линия DC равна BE, угол ADC равен AEB, а~угол ACD — углу ABE. Если из равных линий AD и~AE вычесть равные AB и~AC, остатки BD и~CE также будут равны. А~значит, в~треугольниках BDC и~CEB стороны BD и~DC соответственно равны CE и~EB и~углы D и~E, заключенные между этими сторонами, также равны. А~значит \bycref{prop:I.IV}, углы DBC и~ECB, заключенные между третьей стороной BC и~продолжениями сторон AB и~AC, также равны. Кроме того, углы DCB и~EBC равны, если равные углы вычитаются из углов DCA и~EBA, равенство которых было показано выше, а~значит остатки, то есть углы ABC и~ACB, противолежащие равным сторонам, будут равны. +В треугольниках DAC и~EAB стороны DA и~AC соответственно равны EA и~AB, а~прилежащий угол A общий обоим. Следовательно \byref{prop:I.IV}, линия DC равна BE, угол ADC равен AEB, а~угол ACD — углу ABE. Если из равных линий AD и~AE вычесть равные AB и~AC, остатки BD и~CE также будут равны. А~значит, в~треугольниках BDC и~CEB стороны BD и~DC соответственно равны CE и~EB и~углы D и~E, заключенные между этими сторонами, также равны. А~значит \byref{prop:I.IV}, углы DBC и~ECB, заключенные между третьей стороной BC и~продолжениями сторон AB и~AC, также равны. Кроме того, углы DCB и~EBC равны, если равные углы вычитаются из углов DCA и~EBA, равенство которых было показано выше, а~значит остатки, то есть углы ABC и~ACB, противолежащие равным сторонам, будут равны. \emph{Следовательно, в~равностороннем треугольнике…} и~т. д. @@ -1158,14 +1157,14 @@ \chapter*{Предложения} draw byLabelLineEnd(A, B, 1); draw byLabelLineEnd(B, A, 0); }} -\bycref{post:I.III}. +\byref{post:I.III}. -Проведем \drawUnitLine{CA} и~\drawUnitLine{BC} \bycref{post:I.I}.\\ +Проведем \drawUnitLine{CA} и~\drawUnitLine{BC} \byref{post:I.I}.\\ Тогда \drawLine[bottom][triangleABC]{AB,CA,BC} равносторонний. -Поскольку $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{AB} = \drawUnitLine{BC}$ \bycref{def:I.XV}, -$\therefore \drawUnitLine{CA} = \drawUnitLine{BC}$ \bycref{ax:I.I},\\ +Поскольку $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{def:I.XV}\\ +и $\drawUnitLine{AB} = \drawUnitLine{BC}$ \byref{def:I.XV}, +$\therefore \drawUnitLine{CA} = \drawUnitLine{BC}$ \byref{ax:I.I},\\ и значит, \triangleABC\ и~есть искомый треугольник. \end{center} @@ -1214,14 +1213,14 @@ \chapter*{Предложения} } отложить прямую, равную данной прямой \drawUnitLine{BC}.} \begin{center} -Проведем \drawUnitLine{AB} \bycref{post:I.I}, построим \drawFromCurrentPicture[middle]{ +Проведем \drawUnitLine{AB} \byref{post:I.I}, построим \drawFromCurrentPicture[middle]{ startAutoLabeling; startTempScale(scaleFactor*3); draw byNamedLineSeq(0)(AB,BD,DA); stopTempScale; stopAutoLabeling; -} \bycref{prop:I.I},\\ -продлим \drawUnitLine{BD} \bycref{post:I.II},\\ +} \byref{prop:I.I},\\ +продлим \drawUnitLine{BD} \byref{post:I.II},\\ опишем \drawFromCurrentPicture{ draw byNamedLine (BC); @@ -1229,24 +1228,24 @@ \chapter*{Предложения} draw byLabelLineEnd(B, C, 0); draw byLabelLineEnd(C, B, 0); } -\bycref{post:I.III} и +\byref{post:I.III} и \drawFromCurrentPicture{ draw byNamedLineSeq(0)(BD, BE); draw byNamedCircle(A); draw byLabelLineEnd(D, E, 0); draw byLabelLineEnd(E, D, 1); } -\bycref{post:I.III}. +\byref{post:I.III}. -Продлим \drawUnitLine{DA} \bycref{post:I.II},\\ +Продлим \drawUnitLine{DA} \byref{post:I.II},\\ тогда искомая прямая~— это \drawUnitLine{AF}. -Поскольку $\drawUnitLine{BE,BD} = \drawUnitLine{DA,AF}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{BD} = \drawUnitLine{DA}$ \bycref{\constref},\\ -$\therefore \drawUnitLine{BE} = \drawUnitLine{AF}$ \bycref{ax:I.III},\\ -но \bycref{def:I.XV} $\drawUnitLine{BC} = \drawUnitLine{BE} = \drawUnitLine{AF}$. +Поскольку $\drawUnitLine{BE,BD} = \drawUnitLine{DA,AF}$ \byref{def:I.XV}\\ +и $\drawUnitLine{BD} = \drawUnitLine{DA}$ \byref{\constref},\\ +$\therefore \drawUnitLine{BE} = \drawUnitLine{AF}$ \byref{ax:I.III},\\ +но \byref{def:I.XV} $\drawUnitLine{BC} = \drawUnitLine{BE} = \drawUnitLine{AF}$. -$\therefore \drawUnitLine{AF}$, проведенная из данной точки \pointA, равна данной прямой \drawUnitLine{BC} \bycref{ax:I.I}. +$\therefore \drawUnitLine{AF}$, проведенная из данной точки \pointA, равна данной прямой \drawUnitLine{BC} \byref{ax:I.I}. \end{center} \qed @@ -1279,7 +1278,7 @@ \chapter*{Предложения} \problem{О}{т}{большей \drawUnitLine{AB,BC} из двух данных прямых отнять прямую, равную меньшей \drawUnitLine{EF}.} \begin{center} -Проведем $\drawUnitLine{AD} = \drawUnitLine{EF}$ \bycref{prop:I.II}. +Проведем $\drawUnitLine{AD} = \drawUnitLine{EF}$ \byref{prop:I.II}. Опишем \drawFromCurrentPicture{ @@ -1287,14 +1286,14 @@ \chapter*{Предложения} draw byNamedCircle(A); draw byLabelLineEnd(D, A, 0); draw byLabelLineEnd(A, D, 0); -} \bycref{post:I.III}. +} \byref{post:I.III}. Тогда $\drawUnitLine{EF} = \drawUnitLine{AB}$. -Поскольку $\drawUnitLine{AD} = \drawUnitLine{AB}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{EF} = \drawUnitLine{AD}$ \bycref{\constref}. +Поскольку $\drawUnitLine{AD} = \drawUnitLine{AB}$ \byref{def:I.XV}\\ +и $\drawUnitLine{EF} = \drawUnitLine{AD}$ \byref{\constref}. -$\therefore \drawUnitLine{EF} = \drawUnitLine{AB}$ \bycref{ax:I.I}. +$\therefore \drawUnitLine{EF} = \drawUnitLine{AB}$ \byref{ax:I.I}. \end{center} \qed @@ -1338,7 +1337,7 @@ \chapter*{Предложения} Тогда \drawUnitLine{CA} при наложении совпадет с~\drawUnitLine{FD}. $\therefore$ \drawUnitLine{BC} совпадает с~\drawUnitLine{EF},\\ -или же две прямые будут содержать пространство, что невозможно \bycref{ax:I.X}. +или же две прямые будут содержать пространство, что невозможно \byref{ax:I.X}. $\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$,\\ $\drawAngle{B} = \drawAngle{E}$ и~$\drawAngle{C} = \drawAngle{F}$. @@ -1382,8 +1381,8 @@ \chapter*{Предложения} \problem[3]{У}{глы}{при основании любого равнобедренного треугольника \drawLine[bottom]{BC,AC,AB} равны между собой, и~по продолжении равных сторон углы под основанием будут равны между собой.} \begin{center} -Продлим \drawUnitLine{AB} и~\drawUnitLine{AC} \bycref{post:I.II},\\ -возьмем $\drawUnitLine{BD} = \drawUnitLine{CE}$ \bycref{prop:I.III},\\ +Продлим \drawUnitLine{AB} и~\drawUnitLine{AC} \byref{post:I.II},\\ +возьмем $\drawUnitLine{BD} = \drawUnitLine{CE}$ \byref{prop:I.III},\\ проведем \drawUnitLine{BE} и~\drawUnitLine{CD}. Тогда в @@ -1406,9 +1405,9 @@ \chapter*{Предложения} }\\ получим $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$ (конст.),\\ \drawAngle{BAC} общий обоим,\\ -и $\drawUnitLine{AB} = \drawUnitLine{AC}$ \bycref{\hypref}. +и $\drawUnitLine{AB} = \drawUnitLine{AC}$ \byref{\hypref}. -$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ и~$\drawAngle{CEB} = \drawAngle{BDC}$ \bycref{prop:I.IV}. +$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ и~$\drawAngle{CEB} = \drawAngle{BDC}$ \byref{prop:I.IV}. Так же у~\drawFromCurrentPicture{ startAutoLabeling; @@ -1430,8 +1429,8 @@ \chapter*{Предложения} stopAutoLabeling; }\\ получим $\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{CEB} = \drawAngle{BDC}$ и~$\drawUnitLine{BE} = \drawUnitLine{CD}$,\\ -$\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ и~$\drawAngle{DCB} = \drawAngle{CBE}$ \bycref{prop:I.IV},\\ -но $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$ \bycref{ax:I.III}. +$\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ и~$\drawAngle{DCB} = \drawAngle{CBE}$ \byref{prop:I.IV},\\ +но $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$ \byref{ax:I.III}. \end{center} \qed @@ -1458,15 +1457,15 @@ \chapter*{Предложения} \drawCurrentPictureInMargin \problem[4]{Е}{сли}{у любого треугольника \drawLine[bottom][triangleABD]{CA,CD,BD,AB} два угла \drawAngle{A} и~\drawAngle{B} равны между собой, то и~стороны \drawUnitLine{CA,CD} и~\drawUnitLine{BD}, стягивающие равные углы, будут равны.} -Предположим, что стороны не равны и~одна из них \drawUnitLine{CA,CD} больше, чем другая \drawUnitLine{BD}, тогда отрежем от нее $\drawUnitLine{CA} = \drawUnitLine{BD}$ \bycref{prop:I.III} и~проведем \drawUnitLine{BC}. +Предположим, что стороны не равны и~одна из них \drawUnitLine{CA,CD} больше, чем другая \drawUnitLine{BD}, тогда отрежем от нее $\drawUnitLine{CA} = \drawUnitLine{BD}$ \byref{prop:I.III} и~проведем \drawUnitLine{BC}. \begin{center} Тогда в~\drawLine[bottom]{BC,AB,CA} и~\triangleABD\\ -$\drawUnitLine{CA} = \drawUnitLine{BD}$ \bycref{\constref},\\ -$\drawAngle{A} = \drawAngle{B}$ \bycref{\hypref}\\ +$\drawUnitLine{CA} = \drawUnitLine{BD}$ \byref{\constref},\\ +$\drawAngle{A} = \drawAngle{B}$ \byref{\hypref}\\ и \drawUnitLine{AB} общая обоим. -$\therefore$ эти треугольники равны \bycref{prop:I.IV},\\ +$\therefore$ эти треугольники равны \byref{prop:I.IV},\\ часть равна целому, что невозможно. $\therefore$ ни одна из сторон \drawUnitLine{CA,CD} или \drawUnitLine{BD} не больше другой,\\ @@ -1529,14 +1528,14 @@ \chapter*{Предложения} \drawUnitLine{BC}&=\drawUnitLine{BD}\\ \end{aligned}\right\}$,\\ затем проведем \drawUnitLine{CD}, тогда\\ -$\drawAngle{C,DCB} = \drawAngle{CDA}$ \bycref{prop:I.V}. +$\drawAngle{C,DCB} = \drawAngle{CDA}$ \byref{prop:I.V}. $\therefore\drawAngle{DCB} < \drawAngle{CDA}$. $\left. \begin{aligned} \mbox{И } \therefore\drawAngle{DCB} &< \drawAngle{CDA,D}\mbox{,}\\ -\mbox{но \bycref{prop:I.V}} \drawAngle{DCB} &= \drawAngle{CDA,D}\\ +\mbox{но \byref{prop:I.V}} \drawAngle{DCB} &= \drawAngle{CDA,D}\\ \end{aligned} \right\}\mbox{, что невозможно.}$ \end{center} @@ -1575,7 +1574,7 @@ \chapter*{Предложения} \begin{center} Если совместить равные основания \drawUnitLine{BC} и~\drawUnitLine{EF} так, чтобы треугольники находились по одну сторону,\\ а~их равные стороны \drawUnitLine{AB} и~\drawUnitLine{DE}, \drawUnitLine{CA} и~\drawUnitLine{FD} были смежными,\\ -вершина одного будет совпадать с~вершиной другого \bycref{prop:I.VII}. +вершина одного будет совпадать с~вершиной другого \byref{prop:I.VII}. $\therefore$ стороны \drawUnitLine{AB} и~\drawUnitLine{CA}\\ будут совпадать с~\drawUnitLine{DE} и~\drawUnitLine{FD}\\ @@ -1613,16 +1612,16 @@ \chapter*{Предложения} \problem{Р}{ассечь}{данный прямолинейный угол \drawAngle{BAD,CAD} пополам.} \begin{center} -Возьмем $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{prop:I.III}. +Возьмем $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{prop:I.III}. -Проведем \drawUnitLine{BC}, на которой построим \drawLine{CD,DB,BC} \bycref{prop:I.I},\\ +Проведем \drawUnitLine{BC}, на которой построим \drawLine{CD,DB,BC} \byref{prop:I.I},\\ проведем \drawUnitLine{AD}. Поскольку в \drawLine[middle]{DB,BA,AD} и \drawLine[middle]{CD,DA,AC}\\ -$\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{\constref},\\ -$\drawUnitLine{CD} = \drawUnitLine{DB}$ \bycref{\constref}\\ +$\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{\constref},\\ +$\drawUnitLine{CD} = \drawUnitLine{DB}$ \byref{\constref}\\ и \drawUnitLine{AD} общая обоим,\\ -$\therefore \drawAngle{BAD} = \drawAngle{CAD}$ \bycref{prop:I.VIII}. +$\therefore \drawAngle{BAD} = \drawAngle{CAD}$ \byref{prop:I.VIII}. \end{center} \qed @@ -1649,12 +1648,12 @@ \chapter*{Предложения} \problem{Р}{ассечь}{данную ограниченную прямую линию \drawUnitLine{DB,CD} пополам.} \begin{center} -Построим \drawLine[bottom]{AB,CA,CD,DB} \bycref{prop:I.I},\\ -проведем \drawUnitLine{AD}, делая $\drawAngle{BAD} = \drawAngle{CAD}$ \bycref{prop:I.IX}. +Построим \drawLine[bottom]{AB,CA,CD,DB} \byref{prop:I.I},\\ +проведем \drawUnitLine{AD}, делая $\drawAngle{BAD} = \drawAngle{CAD}$ \byref{prop:I.IX}. -Тогда $\drawUnitLine{BD} = \drawUnitLine{DC}$ \bycref{prop:I.IV},\\ +Тогда $\drawUnitLine{BD} = \drawUnitLine{DC}$ \byref{prop:I.IV},\\ ведь в \drawLine[bottom]{DB,BA,AD} и \drawLine[bottom]{CD,DA,AC}\\ -$\drawUnitLine{AB} = \drawUnitLine{AC}$, $\drawAngle{BAD} = \drawAngle{CAD}$ \bycref{\constref}\\ +$\drawUnitLine{AB} = \drawUnitLine{AC}$, $\drawAngle{BAD} = \drawAngle{CAD}$ \byref{\constref}\\ и \drawUnitLine{AD} общая обоим. Следовательно, данная линия рассечена пополам. @@ -1691,17 +1690,17 @@ \chapter*{Предложения} \begin{center} Возьмем любую точку \drawPointL[middle][CA]{C} на данной прямой,\\ -отсечем $\drawUnitLine{DB} = \drawUnitLine{CD}$ \bycref{prop:I.III},\\ -построим \drawLine[bottom]{AB,CA,CD,DB} \bycref{prop:I.I},\\ +отсечем $\drawUnitLine{DB} = \drawUnitLine{CD}$ \byref{prop:I.III},\\ +построим \drawLine[bottom]{AB,CA,CD,DB} \byref{prop:I.I},\\ проведем \drawUnitLine{AD}, и~она будет перпендикуляром к~данной прямой. Поскольку в \drawLine[bottom]{DB,BA,AD} и \drawLine[bottom]{CD,DA,AC} -$\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{\constref},\\ -$\drawUnitLine{CD} = \drawUnitLine{DB}$ \bycref{\constref}\\ +$\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{\constref},\\ +$\drawUnitLine{CD} = \drawUnitLine{DB}$ \byref{\constref}\\ и \drawUnitLine{AD} общая обоим,\\ -$\therefore \drawAngle{ADB} = \drawAngle{CDA}$ \bycref{prop:I.VIII}. +$\therefore \drawAngle{ADB} = \drawAngle{CDA}$ \byref{prop:I.VIII}. -$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \bycref{def:I.X}. +$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \byref{def:I.X}. \end{center} \qed @@ -1747,17 +1746,17 @@ \chapter*{Предложения} \begin{center} Взяв данную точку \drawPointL{A} в~качестве центра по одну сторону прямой и~любое расстояние, позволяющее достигнуть другой стороны, построим \drawArc{O}. -Возьмем $\drawUnitLine{DB} = \drawUnitLine{CD}$ \bycref{prop:I.X},\\ +Возьмем $\drawUnitLine{DB} = \drawUnitLine{CD}$ \byref{prop:I.X},\\ проведем \drawUnitLine{AB}, \drawUnitLine{CA} и~\drawUnitLine{AD}. Тогда $\drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$. Поскольку в \drawLine[bottom]{DB,BA,AD} и \drawLine[bottom]{CD,DA,AC} -$\drawUnitLine{DB} = \drawUnitLine{CD}$ \bycref{\constref},\\ +$\drawUnitLine{DB} = \drawUnitLine{CD}$ \byref{\constref},\\ \drawUnitLine{AD} общая обоим\\ -и $\drawUnitLine{AB} = \drawUnitLine{CA}$ \bycref{def:I.XV},\\ -$\therefore \drawAngle{ADB} = \drawAngle{CDA}$\bycref{prop:I.VIII},\\ -и $\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \bycref{def:I.X}. +и $\drawUnitLine{AB} = \drawUnitLine{CA}$ \byref{def:I.XV},\\ +$\therefore \drawAngle{ADB} = \drawAngle{CDA}$\byref{prop:I.VIII},\\ +и $\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \byref{def:I.X}. \end{center} \qed @@ -1786,15 +1785,15 @@ \chapter*{Предложения} \begin{center} Если $\drawUnitLine{ED} \perp \drawUnitLine{BC}$, тогда\\ -$\drawAngle{ADB,EDA} + \drawAngle{CDE} = \drawTwoRightAngles$ \bycref{def:I.X}. +$\drawAngle{ADB,EDA} + \drawAngle{CDE} = \drawTwoRightAngles$ \byref{def:I.X}. Но если \drawUnitLine{ED} будет не $\perp$ к~\drawUnitLine{BC},\\ -проведем $\drawUnitLine{AD} \perp \drawUnitLine{BC}$ \bycref{prop:I.XI}. +проведем $\drawUnitLine{AD} \perp \drawUnitLine{BC}$ \byref{prop:I.XI}. -$\drawAngle{ADB} +\drawAngle{CDE,EDA} = \drawTwoRightAngles$ \bycref{\constref},\\ +$\drawAngle{ADB} +\drawAngle{CDE,EDA} = \drawTwoRightAngles$ \byref{\constref},\\ $\drawAngle{ADB} = \drawAngle{CDE,EDA} = \drawAngle{EDA} + \drawAngle{CDE}$. -$\therefore \drawAngle{ADB} + \drawAngle{CDE,EDA} = \drawAngle{ADB} + \drawAngle{EDA} + \drawAngle{CDE}$ \bycref{ax:I.II}\\ +$\therefore \drawAngle{ADB} + \drawAngle{CDE,EDA} = \drawAngle{ADB} + \drawAngle{EDA} + \drawAngle{CDE}$ \byref{ax:I.II}\\ $= \drawAngle{ADB,EDA} + \drawAngle{CDE} = \drawTwoRightAngles$. \end{center} @@ -1829,8 +1828,8 @@ \chapter*{Предложения} Но, согласно гипотезе, $\drawAngle{BDA} + \drawAngle{CDA} = \drawTwoRightAngles$ -$\therefore\drawAngle{CDA,EDC} = \drawAngle{CDA}$ \bycref{ax:I.III},\\ -что невозможно \bycref{ax:I.IX}. +$\therefore\drawAngle{CDA,EDC} = \drawAngle{CDA}$ \byref{ax:I.III},\\ +что невозможно \byref{ax:I.IX}. $\therefore \drawUnitLine{ED}$ не является продолжением \drawUnitLine{BD}, и~то же можно показать для любой другой прямой линии, за исключением \drawUnitLine{DC}. @@ -1862,11 +1861,11 @@ \chapter*{Предложения} \problem{Е}{сли}{две прямых линии \drawUnitLine{AB} и~\drawUnitLine{CD} пересекаются, вертикальные углы \drawAngle{BEC} и~\drawAngle{AED}, \drawAngle{CEA} и~\drawAngle{DEB} будут равны между собой.} \begin{center} -$\drawAngle{BEC} + \drawAngle{CEA} = \drawTwoRightAngles$ \bycref{prop:I.XIII}. +$\drawAngle{BEC} + \drawAngle{CEA} = \drawTwoRightAngles$ \byref{prop:I.XIII}. -$\drawAngle{AED} + \drawAngle{CEA} = \drawTwoRightAngles$\bycref{prop:I.XIII}. +$\drawAngle{AED} + \drawAngle{CEA} = \drawTwoRightAngles$\byref{prop:I.XIII}. -$\therefore \drawAngle{BEC} = \drawAngle{AED}$ \bycref{ax:I.III}. +$\therefore \drawAngle{BEC} = \drawAngle{AED}$ \byref{ax:I.III}. Таким же образом можно показать,\\ что $\drawAngle{CEA} = \drawAngle{DEB}$. @@ -1911,15 +1910,15 @@ \chapter*{Предложения} } \begin{center} -Сделаем $\drawUnitLine{BE} = \drawUnitLine{EC}$ \bycref{prop:I.X},\\ +Сделаем $\drawUnitLine{BE} = \drawUnitLine{EC}$ \byref{prop:I.X},\\ проведем \drawUnitLine{AE} и~продлим до $\drawUnitLine{ED} = \drawUnitLine{AE}$,\\ проведем \drawUnitLine{CD}. В~\drawLine{BE,AE,AB} и~\drawLine{EC,ED,CD}\\ -$\drawUnitLine{BE} = \drawUnitLine{EC}$, $\drawAngle{AEB} = \drawAngle{DEC}$ \bycref{prop:I.XV}\\ -и~$\drawUnitLine{AE} = \drawUnitLine{ED}$ \bycref{\constref}. +$\drawUnitLine{BE} = \drawUnitLine{EC}$, $\drawAngle{AEB} = \drawAngle{DEC}$ \byref{prop:I.XV}\\ +и~$\drawUnitLine{AE} = \drawUnitLine{ED}$ \byref{\constref}. -$\therefore \drawAngle{B} = \drawAngle{ECD}$ \bycref{prop:I.IV},\\ +$\therefore \drawAngle{B} = \drawAngle{ECD}$ \byref{prop:I.IV},\\ $\therefore \drawAngleWithSides{FCE} > \drawAngle{B}$. Так же можно показать, что при продлении \drawUnitLine{BC} $\drawAngle{GCA} > \drawAngle{A}$, \\ @@ -1954,7 +1953,7 @@ \chapter*{Предложения} Продлим \drawUnitLine{AC}, тогда\\ $\drawAngle{ACB} + \drawAngle{BCD} = \drawTwoRightAngles$. -Но $\drawAngle{BCD} > \drawAngle{A}$ \bycref{prop:I.XVI}. +Но $\drawAngle{BCD} > \drawAngle{A}$ \byref{prop:I.XVI}. $\therefore \drawAngle{ACB} + \drawAngle{A} < \drawTwoRightAngles$. \end{center} @@ -1992,11 +1991,11 @@ \chapter*{Предложения} \problem[3]{В}{о}{всяком треугольнике \drawLine{DC,BC,BA,AD} если одна сторона \drawUnitLine{AD,DC} больше другой \drawUnitLine[0.5cm]{BC}, то противолежащий большей стороне угол будет больше противолежащего меньшей стороне угла, т.~е. $\drawAngle{DBC,ABD} > \drawAngle{A}$.} \begin{center} -Сделаем $\drawUnitLine{DC} = \drawUnitLine{BC}$ \bycref{prop:I.III},\\ проведем \drawUnitLine{DB}. +Сделаем $\drawUnitLine{DC} = \drawUnitLine{BC}$ \byref{prop:I.III},\\ проведем \drawUnitLine{DB}. -Тогда $\drawAngle{D} = \drawAngle{DBC}$ \bycref{prop:I.V}. +Тогда $\drawAngle{D} = \drawAngle{DBC}$ \byref{prop:I.V}. -Но $\drawAngle{D} > \drawAngle{A}$ \bycref{prop:I.XVI}. +Но $\drawAngle{D} > \drawAngle{A}$ \byref{prop:I.XVI}. $\therefore \drawAngle{DBC} > \drawAngle{A}$,\\ и тем более $\drawAngle{DBC,ABD} > \drawAngle{A}$. @@ -2027,12 +2026,12 @@ \chapter*{Предложения} $\drawUnitLine{BC} =$ или $< \drawUnitLine{CA}$. Если $\drawUnitLine{BC} = \drawUnitLine{CA}$, тогда\\ -$\drawAngle{A} = \drawAngle{B}$ \bycref{prop:I.V},\\ +$\drawAngle{A} = \drawAngle{B}$ \byref{prop:I.V},\\ что противоречит гипотезе. \drawUnitLine{BC} также не меньше \drawUnitLine{CA},\\ поскольку если $\drawUnitLine{BC} < \drawUnitLine{CA}$,\\ -то $\drawAngle{A} < \drawAngle{B}$ \bycref{prop:I.XVIII}\\ +то $\drawAngle{A} < \drawAngle{B}$ \byref{prop:I.XVIII}\\ что противоречит гипотезе. $\therefore \drawUnitLine{BC} > \drawUnitLine{CA}$. @@ -2065,16 +2064,16 @@ \chapter*{Предложения} \begin{center} Продлим \drawUnitLine{DA}\\ -и сделаем $\drawUnitLine{CD} = \drawUnitLine{BD}$ \bycref{prop:I.III}. +и сделаем $\drawUnitLine{CD} = \drawUnitLine{BD}$ \byref{prop:I.III}. Проведем \drawUnitLine{BC}. -Тогда, поскольку $\drawUnitLine{CD} = \drawUnitLine{BD}$ \bycref{\constref},\\ -$\drawAngle{CBD} = \drawAngle{C}$ \bycref{prop:I.V}. +Тогда, поскольку $\drawUnitLine{CD} = \drawUnitLine{BD}$ \byref{\constref},\\ +$\drawAngle{CBD} = \drawAngle{C}$ \byref{prop:I.V}. -$\therefore \drawAngle{CBD,DBA} > \drawAngle{C}$ \bycref{ax:I.IX}. +$\therefore \drawAngle{CBD,DBA} > \drawAngle{C}$ \byref{ax:I.IX}. -$\therefore \drawUnitLine{DA} + \drawUnitLine{CD} > \drawUnitLine{AB}$ \bycref{prop:I.XIX}. +$\therefore \drawUnitLine{DA} + \drawUnitLine{CD} > \drawUnitLine{AB}$ \byref{prop:I.XIX}. И $\therefore \drawUnitLine{DA} + \drawUnitLine{BD} > \drawUnitLine{AB}$. \end{center} @@ -2111,17 +2110,17 @@ \chapter*{Предложения} \begin{center} Продлим \drawSizedLine{AD},\\ -$\drawSizedLine{CA} + \drawSizedLine{EC} > \drawSizedLine{AD,DE}$ \bycref{prop:I.XX},\\ +$\drawSizedLine{CA} + \drawSizedLine{EC} > \drawSizedLine{AD,DE}$ \byref{prop:I.XX},\\ добавим к~каждой \drawSizedLine{BE},\\ -$\drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD,DE} + \drawSizedLine{BE}$ \bycref{ax:I.IV}. +$\drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD,DE} + \drawSizedLine{BE}$ \byref{ax:I.IV}. Таким же образом можно показать, что\\ $\drawSizedLine{AD,DE} + \drawSizedLine{BE} > \drawSizedLine{AD} + \drawSizedLine{BD}$,\\ $\therefore \drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD} + \drawSizedLine{BD}$,\\ что и~требовалось доказать. -Далее $\drawAngle{E} > \drawAngle{C}$ \bycref{prop:I.XVI}\\ -и так же $\drawAngle{D} > \drawAngle{E}$ \bycref{prop:I.XVI}. +Далее $\drawAngle{E} > \drawAngle{C}$ \byref{prop:I.XVI}\\ +и так же $\drawAngle{D} > \drawAngle{E}$ \byref{prop:I.XVI}. $\therefore \drawAngle{D} > \drawAngle{C}$. \end{center} @@ -2177,14 +2176,14 @@ \chapter*{Предложения} таких, что любые две вместе длиннее третьей, составить треугольник.} \begin{center} -Предположим, $\drawSizedLine{AB} = \drawSizedLine{L'}$ \bycref{prop:I.III}. +Предположим, $\drawSizedLine{AB} = \drawSizedLine{L'}$ \byref{prop:I.III}. $\left. \begin{aligned} \mbox{Проведем } \drawSizedLine{BE} &= \drawSizedLine{L''}\\ \mbox{и } \drawSizedLine{AD} &= \drawSizedLine{L'''}\\ \end{aligned} -\right\}\mbox{\bycref{prop:I.II}.}$ +\right\}\mbox{\byref{prop:I.II}.}$ Взяв \drawSizedLine{AD} и~\drawSizedLine{BE} как радиусы, опишем \drawFromCurrentPicture{ @@ -2196,7 +2195,7 @@ \chapter*{Предложения} draw byNamedLine(BE); draw byNamedCircle(B); draw byLabelLineEnd(B, E, 0); draw byLabelLineEnd(E, B, 0); -}} \bycref{post:I.III}. +}} \byref{post:I.III}. Проведем \drawSizedLine{CA} и~\drawSizedLine{BC}. @@ -2208,7 +2207,7 @@ \chapter*{Предложения} \drawSizedLine{BC} &= \drawSizedLine{BE} = \drawSizedLine{L''} \\ \mbox{и } \drawSizedLine{CA} &= \drawSizedLine{AD} = \drawSizedLine{L'''} \\ \end{aligned} -\right\}\mbox{\bycref{\constref}.}$ +\right\}\mbox{\byref{\constref}.}$ \end{center} \qed @@ -2249,12 +2248,12 @@ \chapter*{Предложения} Проведем \drawUnitLine{BC} между любыми двумя точками на сторонах данного угла. \begin{center} -Построим \drawLine[bottom]{HF,GH,FG} \bycref{prop:I.XXII} такой,\\ +Построим \drawLine[bottom]{HF,GH,FG} \byref{prop:I.XXII} такой,\\ что $\drawUnitLine{FG} = \drawUnitLine{AB}$,\\ $\drawUnitLine{HF} = \drawUnitLine{CA}$\\ и $\drawUnitLine{GH} = \drawUnitLine{BC}$. -Тогда $\drawAngle{A} = \drawAngle{F}$ \bycref{prop:I.VIII}. +Тогда $\drawAngle{A} = \drawAngle{F}$ \byref{prop:I.VIII}. \end{center} \qed @@ -2296,18 +2295,18 @@ \chapter*{Предложения} \problem{Е}{сли}{у двух треугольников по две стороны соответственно равны друг другу ($\drawUnitLine{AB} = \drawUnitLine{EF}$ и~$\drawUnitLine{AD} = \drawUnitLine{GE}$) и~угол, заключенный между ними в~одном \drawAngleWithSides{DAB}, больше, чем в~другом \drawAngleWithSides{FEG}, то сторона \drawUnitLine{DB}, противолежащая большему углу, больше стороны, противолежащей меньшему \drawUnitLine{FG}.} \begin{center} -Сделаем $\drawAngleWithSides{BAC} = \drawAngleWithSides{FEG}$ \bycref{prop:I.XXIII}\\ -и $\drawUnitLine{CA} = \drawUnitLine{GE}$ \bycref{prop:I.III},\\ +Сделаем $\drawAngleWithSides{BAC} = \drawAngleWithSides{FEG}$ \byref{prop:I.XXIII}\\ +и $\drawUnitLine{CA} = \drawUnitLine{GE}$ \byref{prop:I.III},\\ проведем \drawUnitLine{CD} и~\drawUnitLine{BC}. -Поскольку $\drawUnitLine{CA} = \drawUnitLine{AD}$ \bycref{ax:I.I,\hypref,\constref}\\ -$\therefore \drawAngle{BDA,CDB} = \drawAngle{DCA}$ \bycref{prop:I.V}, +Поскольку $\drawUnitLine{CA} = \drawUnitLine{AD}$ \byref{ax:I.I,\hypref,\constref}\\ +$\therefore \drawAngle{BDA,CDB} = \drawAngle{DCA}$ \byref{prop:I.V}, но $\drawAngle{CDB} < \drawAngle{DCA}$,\\ и $\therefore \drawAngle{CDB} < \drawAngle{DCA,ACB}$. -$\therefore \drawUnitLine{DB} > \drawUnitLine{BC}$ \bycref{prop:I.XIX}. +$\therefore \drawUnitLine{DB} > \drawUnitLine{BC}$ \byref{prop:I.XIX}. -Но $\drawUnitLine{BC} = \drawUnitLine{FG}$ \bycref{prop:I.IV}. +Но $\drawUnitLine{BC} = \drawUnitLine{FG}$ \byref{prop:I.IV}. $\therefore \drawUnitLine{DB} > \drawUnitLine{FG}$. \end{center} @@ -2346,12 +2345,12 @@ \chapter*{Предложения} \drawAngle{A} не равен \drawAngle{D},\\ поскольку если $\drawAngle{A} = \drawAngle{D}$,\\ -то $\drawUnitLine{CB} = \drawUnitLine{FE}$ \bycref{prop:I.IV},\\ +то $\drawUnitLine{CB} = \drawUnitLine{FE}$ \byref{prop:I.IV},\\ что противоречит гипотезе. \drawAngle{A} не меньше \drawAngle{D},\\ поскольку если $\drawAngle{A} < \drawAngle{D}$,\\ -то $\drawUnitLine{CB} < \drawUnitLine{FE}$ \bycref{prop:I.XXIV},\\ +то $\drawUnitLine{CB} < \drawUnitLine{FE}$ \byref{prop:I.XXIV},\\ что противоречит гипотезе. $\therefore \drawAngle{A} > \drawAngle{D}$. @@ -2406,11 +2405,11 @@ \chapter*{Предложения} \drawLine[bottom]{GD,EG,DE} получим \\ $\drawUnitLine{CA} = \drawUnitLine{GD}$, $\drawAngle{A} = \drawAngle{D}$, $\drawUnitLine{AB} = \drawUnitLine{DE}$;\\ $\therefore \drawAngle{B} = \drawAngle{GED}$ (pr. 4.)\\ -но $\drawAngle{B} = \drawAngle{GED,FEG}$ \bycref{\hypref}. +но $\drawAngle{B} = \drawAngle{GED,FEG}$ \byref{\hypref}. И следовательно $\drawAngle{GED} = \drawAngle{GED,FEG}$, что не~имеет смысла, а~значит ни \drawUnitLine{CA}, ни \drawUnitLine{GD,FG} не~больше другой, и~$\therefore$ они равны. -$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$, и~$\drawAngle{C} = \drawAngle{F}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$, и~$\drawAngle{C} = \drawAngle{F}$ \byref{prop:I.IV}. \end{center} \vfill\pagebreak @@ -2453,12 +2452,12 @@ \chapter*{Предложения} Тогда в~\drawLine[bottom]{CA,BC,AB} и~\drawLine[bottom]{FD,FG,DG} получим $\drawUnitLine{CA} = \drawUnitLine{FD}$, $\drawUnitLine{AB} = \drawUnitLine{DG}$ и~$\drawAngle{A} = \drawAngle{D}$. -$\therefore \drawAngle{B} = \drawAngle{G}$ \bycref{prop:I.IV},\\ -но $\drawAngle{B} = \drawAngle{E}$ \bycref{\hypref}. +$\therefore \drawAngle{B} = \drawAngle{G}$ \byref{prop:I.IV},\\ +но $\drawAngle{B} = \drawAngle{E}$ \byref{\hypref}. -$\therefore \drawAngle{G} = \drawAngle{E}$, что не~имеет смысла \bycref{prop:I.XVI}. % question: при чем тут I.XVI ? +$\therefore \drawAngle{G} = \drawAngle{E}$, что не~имеет смысла \byref{prop:I.XVI}. % question: при чем тут I.XVI ? -Следовательно, ни \drawUnitLine{AB}, ни \drawUnitLine{DG,GE} не~больше другой, а~значит, они равны. Следовательно (согласно \bycref{prop:I.IV}), треугольники равны во всех отношениях. +Следовательно, ни \drawUnitLine{AB}, ни \drawUnitLine{DG,GE} не~больше другой, а~значит, они равны. Следовательно (согласно \byref{prop:I.IV}), треугольники равны во всех отношениях. \end{center} \qed @@ -2496,7 +2495,7 @@ \chapter*{Предложения} Если \drawUnitLine{CD} не параллельна \drawUnitLine{AB}, то они сойдутся по продолжении. -Если это возможно, пусть они не будут параллельны, но сойдутся, если их продолжить; тогда внешний угол \drawAngle{HEB} будет больше \drawAngle{CFG} \bycref{prop:I.XVI}, но они равны \bycref{\hypref}, что невозможно. Таким же образом можно показать, что они не сойдутся с~другой стороны, $\therefore$ они параллельны. +Если это возможно, пусть они не будут параллельны, но сойдутся, если их продолжить; тогда внешний угол \drawAngle{HEB} будет больше \drawAngle{CFG} \byref{prop:I.XVI}, но они равны \byref{\hypref}, что невозможно. Таким же образом можно показать, что они не сойдутся с~другой стороны, $\therefore$ они параллельны. \qed @@ -2531,16 +2530,16 @@ \chapter*{Предложения} \begin{center} Во-первых, если $\drawAngle{GEA} = \drawAngle{CFG}$,\\ -то $\drawAngle{GEA} = \drawAngle{HEB}$ \bycref{prop:I.XV},\\ -$\therefore \drawAngle{CFG} = \drawAngle{HEB} \therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXVII}. +то $\drawAngle{GEA} = \drawAngle{HEB}$ \byref{prop:I.XV},\\ +$\therefore \drawAngle{CFG} = \drawAngle{HEB} \therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXVII}. Во-вторых, если $\drawAngle{CFG} + \drawAngle{AEH} = \drawTwoRightAngles$,\\ -то $\drawAngle{AEH} + \drawAngle{HEB} = \drawTwoRightAngles$ \bycref{prop:I.XIII},\\ -$\therefore \drawAngle{CFG} + \drawAngle{AEH} = \drawAngle{AEH} + \drawAngle{HEB}$ \bycref{ax:I.I}. +то $\drawAngle{AEH} + \drawAngle{HEB} = \drawTwoRightAngles$ \byref{prop:I.XIII},\\ +$\therefore \drawAngle{CFG} + \drawAngle{AEH} = \drawAngle{AEH} + \drawAngle{HEB}$ \byref{ax:I.I}. -$\therefore \drawAngle{CFG} = \drawAngle{HEB}$ \bycref{ax:I.III}. +$\therefore \drawAngle{CFG} = \drawAngle{HEB}$ \byref{ax:I.III}. -$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXVII}. +$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXVII}. \end{center} \qed @@ -2581,17 +2580,17 @@ \chapter*{Предложения} \problem{П}{рямая}{\drawUnitLine{GH}, падающая на две параллельные прямые \drawUnitLine{AB} и~\drawUnitLine{CD}, образует накрест лежащие углы, равные между собой, внешний и~противолежащий с~той же стороны внутренний углы, равные между собой, а~также внутренние односторонние углы, равные двум прямым углам.} \begin{center} -Если накрест лежащие углы \drawAngle{IEA,HEI} и~\drawAngle{GFD} не равны,\\ проведем \drawUnitLine{IE} так, чтобы $\drawAngle{HEI} = \drawAngle{GFD}$ \bycref{prop:I.XXIII}. +Если накрест лежащие углы \drawAngle{IEA,HEI} и~\drawAngle{GFD} не равны,\\ проведем \drawUnitLine{IE} так, чтобы $\drawAngle{HEI} = \drawAngle{GFD}$ \byref{prop:I.XXIII}. -$\therefore \drawUnitLine{IE,EJ} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXVII}\\ -и $\therefore$ две пересекающихся прямых параллельны одной и~той же прямой, что невозможно \bycref{ax:I.XII}. +$\therefore \drawUnitLine{IE,EJ} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXVII}\\ +и $\therefore$ две пересекающихся прямых параллельны одной и~той же прямой, что невозможно \byref{ax:I.XII}. А значит, \drawAngle{IEA,HEI} и~\drawAngle{GFD} не являются неравными,\\ -то есть они равны $\drawAngle{IEA,HEI} = \drawAngle{GEB}$ \bycref{prop:I.XV}. +то есть они равны $\drawAngle{IEA,HEI} = \drawAngle{GEB}$ \byref{prop:I.XV}. $\therefore \drawAngle{GEB} = \drawAngle{GFD}$, внешний угол равен внутреннему, противолежащему с~той же стороны:\\ если к~обоим добавить \drawAngle{BEH},\\ -то $\drawAngle{GFD} + \drawAngle{BEH} = \drawAngle{BEH,GEB} = \drawTwoRightAngles$ \bycref{prop:I.XIII}. +то $\drawAngle{GFD} + \drawAngle{BEH} = \drawAngle{BEH,GEB} = \drawTwoRightAngles$ \byref{prop:I.XIII}. \end{center} Другими словами, два внутренних угла по одну сторону пересекающей прямой равны двум прямым углам. @@ -2632,11 +2631,11 @@ \chapter*{Предложения} \begin{center} Пусть \drawUnitLine{JK} пересекает $\left\{\vcenter{\nointerlineskip\hbox{\drawUnitLine{AB}}\nointerlineskip\hbox{\drawUnitLine{CD}}\nointerlineskip\hbox{\drawUnitLine{EF}}}\right\}$. -Тогда $\drawAngle{G} = \drawAngle{I} = \drawAngle{H}$ \bycref{prop:I.XXIX}. +Тогда $\drawAngle{G} = \drawAngle{I} = \drawAngle{H}$ \byref{prop:I.XXIX}. $\therefore \drawAngle{G} = \drawAngle{H}$. -$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{EF}$ \bycref{prop:I.XXVIII}. +$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{EF}$ \byref{prop:I.XXVIII}. \end{center} \qed @@ -2669,9 +2668,9 @@ \chapter*{Предложения} \begin{center} Проведем \drawUnitLine{EF} из точки \drawPointL[middle][EB]{E} к~любой точке \drawPointL[middle][CF]{F} на \drawUnitLine{CF,FD}. -Сделаем $\drawAngle{E} = \drawAngle{F}$ \bycref{prop:I.XXIII}. +Сделаем $\drawAngle{E} = \drawAngle{F}$ \byref{prop:I.XXIII}. -Тогда $\drawUnitLine{AE,EB} \parallel \drawUnitLine{CF,FD}$ \bycref{prop:I.XXVII}. +Тогда $\drawUnitLine{AE,EB} \parallel \drawUnitLine{CF,FD}$ \byref{prop:I.XXVII}. \end{center} \qed @@ -2703,18 +2702,18 @@ \chapter*{Предложения} \begin{center} Через точку \drawPointL[middle][AD,AE]{A} проведем \\ -$\drawUnitLine{AE} \parallel \drawUnitLine{BC}$ \bycref{prop:I.XXXI}. +$\drawUnitLine{AE} \parallel \drawUnitLine{BC}$ \byref{prop:I.XXXI}. Тогда $\left\{ \begin{aligned} \drawAngle{DAE} &= \drawAngle{B}\\ \drawAngle{EAC} &= \drawAngle{C}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXIX}. +\right\}$ \byref{prop:I.XXIX}. -$\therefore \drawAngle{B} + \drawAngle{C} = \drawAngle{DAE,EAC}$ \bycref{ax:I.II}. +$\therefore \drawAngle{B} + \drawAngle{C} = \drawAngle{DAE,EAC}$ \byref{ax:I.II}. -$\therefore \drawAngle{B} + \drawAngle{CAB} + \drawAngle{C} = \drawAngle{DAE,EAC,CAB} = \drawTwoRightAngles$ \bycref{prop:I.XIII}. +$\therefore \drawAngle{B} + \drawAngle{CAB} + \drawAngle{C} = \drawAngle{DAE,EAC,CAB} = \drawTwoRightAngles$ \byref{prop:I.XIII}. \end{center} \qed @@ -2748,13 +2747,13 @@ \chapter*{Предложения} Проведем диагональ \drawUnitLine{AD}. Поскольку в \drawLine{DC,CA,AD} и \drawLine{DA,AB,BD}\\ -$\drawUnitLine{CD} = \drawUnitLine{AB}$ \bycref{\hypref},\\ -$\drawAngle{CDA} = \drawAngle{BAD}$ \bycref{prop:I.XXIX}\\ +$\drawUnitLine{CD} = \drawUnitLine{AB}$ \byref{\hypref},\\ +$\drawAngle{CDA} = \drawAngle{BAD}$ \byref{prop:I.XXIX}\\ и \drawUnitLine{AD} общая обоим,\\ $\therefore \drawUnitLine{AC} = \drawUnitLine{BD}$\\ -и~$\drawAngle{DAC} = \drawAngle{ADB}$ \bycref{prop:I.IV}. +и~$\drawAngle{DAC} = \drawAngle{ADB}$ \byref{prop:I.IV}. -И $\therefore \drawUnitLine{AC} \parallel \drawUnitLine{BD}$ \bycref{prop:I.XXVII}. +И $\therefore \drawUnitLine{AC} \parallel \drawUnitLine{BD}$ \byref{prop:I.XXVII}. \end{center} \qed @@ -2792,7 +2791,7 @@ \chapter*{Предложения} \drawAngle{BAD} &= \drawAngle{CDA}\\ \drawAngle{DAC} &= \drawAngle{ADB}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXIX}\\ +\right\}$ \byref{prop:I.XXIX}\\ и \drawUnitLine{AD} общая для \drawLine{AD,CD,AC} и~\drawLine{AB,BD,AD}. $\therefore \left\{ @@ -2801,11 +2800,11 @@ \chapter*{Предложения} \drawUnitLine{AC} &= \drawUnitLine{BD}\\ \drawAngle{B} &= \drawAngle{C}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXVI}\\ -и $\drawAngle{BAD,DAC} = \drawAngle{CDA,ADB}$ \bycref{ax:I.II}. +\right\}$ \byref{prop:I.XXVI}\\ +и $\drawAngle{BAD,DAC} = \drawAngle{CDA,ADB}$ \byref{ax:I.II}. \end{center} -Следовательно, противоположные стороны и~углы параллелограмма равны. И, поскольку треугольники \drawLine{AD,CD,AC} и~\drawLine{AB,BD,AD} равны во всех отношениях \bycref{prop:I.IV}, диагональ делит параллелограмм на две равные части. +Следовательно, противоположные стороны и~углы параллелограмма равны. И, поскольку треугольники \drawLine{AD,CD,AC} и~\drawLine{AB,BD,AD} равны во всех отношениях \byref{prop:I.IV}, диагональ делит параллелограмм на две равные части. \qed @@ -2849,9 +2848,9 @@ \chapter*{Предложения} \end{aligned} \right\} \begin{aligned} -&\mbox{\bycref{prop:I.XXIX}}\\ -&\mbox{\bycref{prop:I.XXIX}}\\ -&\mbox{\bycref{prop:I.XXXIV}}\\ +&\mbox{\byref{prop:I.XXIX}}\\ +&\mbox{\byref{prop:I.XXIX}}\\ +&\mbox{\byref{prop:I.XXXIV}}\\ \end{aligned}$ $\therefore \drawFromCurrentPicture[middle][polygonABC]{ @@ -2866,7 +2865,7 @@ \chapter*{Предложения} draw byNamedPolygon (EFDG, BEG); stopAutoLabeling; draw byNamedLine (BD); -}$ \bycref{prop:I.XXVI}. +}$ \byref{prop:I.XXVI}. Но $\drawFromCurrentPicture[middle][polygonAFDC]{ startAutoLabeling; @@ -2921,19 +2920,19 @@ \chapter*{Предложения} \begin{center} Проведем \drawUnitLine{CE} и~\drawUnitLine{DF},\\ -$\drawUnitLine{CD} = \drawUnitLine{GH} = \drawUnitLine{EF}$ \bycref{\hypref,prop:I.XXXIV}. +$\drawUnitLine{CD} = \drawUnitLine{GH} = \drawUnitLine{EF}$ \byref{\hypref,prop:I.XXXIV}. $\therefore \drawUnitLine{CD} = \mbox{ и~} \parallel \drawUnitLine{EF}$. -$\therefore \drawUnitLine{CE} = \mbox{ и~} \parallel \drawUnitLine{DF}$ \bycref{prop:I.XXXIII}. +$\therefore \drawUnitLine{CE} = \mbox{ и~} \parallel \drawUnitLine{DF}$ \byref{prop:I.XXXIII}. Следовательно, \drawPolygon[bottom][polygonCDFE]{CDI, IDJE, EFJ} параллелограмм. -Но $\polygonABDC = \polygonCDFE = \polygonEFHG$ \bycref{prop:I.XXXV}. +Но $\polygonABDC = \polygonCDFE = \polygonEFHG$ \byref{prop:I.XXXV}. -$\therefore \polygonABDC = \polygonEFHG$ \bycref{ax:I.I}. +$\therefore \polygonABDC = \polygonEFHG$ \byref{ax:I.I}. \end{center} \qed @@ -2979,21 +2978,21 @@ \chapter*{Предложения} \mbox{Проведем} \drawUnitLine{AC} &\parallel \drawUnitLine{BD}\\ \drawUnitLine{FD} &\parallel \drawUnitLine{EC}\\ \end{aligned} -\right\}\mbox{\bycref{prop:I.XXXI}.}$ +\right\}\mbox{\byref{prop:I.XXXI}.}$ Проведем \drawUnitLine{HI}. \drawPolygon[bottom][polygonABDC]{ABC, BCG, CDG} и \drawPolygon[bottom][polygonEFDC]{DGE, CDG, EFD} -являются параллелограммами на одном основании и~между одними параллельными прямыми и,~следовательно, равны между собой. \bycref{prop:I.XXXV} +являются параллелограммами на одном основании и~между одними параллельными прямыми и,~следовательно, равны между собой. \byref{prop:I.XXXV} $\therefore \left\{ \begin{aligned} \polygonABDC &= \mbox{ дважды } \polygonBCD\\ \polygonEFDC &= \mbox{ дважды } \polygonCDE\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXIV}. +\right\}$ \byref{prop:I.XXXIV}. $\therefore \polygonBCD = \polygonCDE$. \end{center} @@ -3042,14 +3041,14 @@ \chapter*{Предложения} \mbox{Проведем } \drawUnitLine{AC} &\parallel \drawUnitLine{BD}\\ \mbox{и } \drawUnitLine{FH} &\parallel \drawUnitLine{EG}\\ \end{aligned} -\right\}\mbox{\bycref{prop:I.XXXI}.}$ +\right\}\mbox{\byref{prop:I.XXXI}.}$ -$\drawPolygon[bottom][polygonABDC]{ABC, BCD} = \drawPolygon[bottom][polygonEFHG]{EFH, EGH}$ \bycref{prop:I.XXXVI}. +$\drawPolygon[bottom][polygonABDC]{ABC, BCD} = \drawPolygon[bottom][polygonEFHG]{EFH, EGH}$ \byref{prop:I.XXXVI}. -Но $\polygonABDC = \mbox{ дважды } \polygonBCD$ \bycref{prop:I.XXXIV}\\ -и $\polygonEFHG = \mbox{ дважды } \polygonEGH$ \bycref{prop:I.XXXIV}. +Но $\polygonABDC = \mbox{ дважды } \polygonBCD$ \byref{prop:I.XXXIV}\\ +и $\polygonEFHG = \mbox{ дважды } \polygonEGH$ \byref{prop:I.XXXIV}. -$\therefore \polygonBCD = \polygonEGH$ \bycref{ax:I.VII}. +$\therefore \polygonBCD = \polygonEGH$ \byref{ax:I.VII}. \end{center} \qed @@ -3084,15 +3083,15 @@ \chapter*{Предложения} \drawPolygon[bottom][polygonADC]{AEC, ECD} и~\drawPolygon[bottom][polygonBDC]{BED, ECD}, находящиеся на одном основании \drawUnitLine{CD} и~с одной и~той же стороны, находятся между теми же параллельными прямыми.} \begin{center} -Если \drawUnitLine{AB}, соединяющая вершины треугольников, не $\parallel \drawUnitLine{CD}$, проведем $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \bycref{prop:I.XXXI}, касающуюся \drawUnitLine{CG}. +Если \drawUnitLine{AB}, соединяющая вершины треугольников, не $\parallel \drawUnitLine{CD}$, проведем $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \byref{prop:I.XXXI}, касающуюся \drawUnitLine{CG}. Проведем \drawUnitLine{DF}. -Поскольку $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \bycref{\constref}\\ +Поскольку $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \byref{\constref}\\ $\polygonADC = -\drawPolygon[bottom][polygonFDC]{BED, ECD, FBD}$ \bycref{prop:I.XXXVII}. +\drawPolygon[bottom][polygonFDC]{BED, ECD, FBD}$ \byref{prop:I.XXXVII}. -Но $\polygonADC = \polygonBDC$ \bycref{\hypref}. +Но $\polygonADC = \polygonBDC$ \byref{\hypref}. $\therefore \polygonBDC = \polygonFDC$, часть равна целому, что невозможно. @@ -3149,12 +3148,12 @@ \chapter*{Предложения} \begin{center} Если \drawSizedLine{AB}, соединяющая вершины треугольников, не $\parallel \drawSizedLine{CD,DE,EF}$,\\ -проведем \drawSizedLine{AG} $\parallel \drawSizedLine{CD,DE,EF}$ \bycref{prop:I.XXXI},\\ +проведем \drawSizedLine{AG} $\parallel \drawSizedLine{CD,DE,EF}$ \byref{prop:I.XXXI},\\ касающуюся \drawSizedLine{EH}. Проведем \drawSizedLine{FG}. -Поскольку $\drawSizedLine{AG} \parallel \drawSizedLine{CD,DE,EF}$ \bycref{\constref},\\ +Поскольку $\drawSizedLine{AG} \parallel \drawSizedLine{CD,DE,EF}$ \byref{\constref},\\ $\polygonACD = \drawFromCurrentPicture[bottom][polygonGEF]{ startAutoLabeling; @@ -3201,10 +3200,10 @@ \chapter*{Предложения} \begin{center} Проведем диагональ \drawUnitLine{AE}. -Тогда $\drawPolygon[bottom][polygonAED]{AGD,DEG} = \polygonCED$ \bycref{prop:I.XXXVII},\\ -$\polygonABED = \mbox{ дважды } \polygonAED$ \bycref{prop:I.XXXIV}. +Тогда $\drawPolygon[bottom][polygonAED]{AGD,DEG} = \polygonCED$ \byref{prop:I.XXXVII},\\ +$\polygonABED = \mbox{ дважды } \polygonAED$ \byref{prop:I.XXXIV}. -$\therefore \polygonABED = \mbox{ дважды } \polygonCED$ \bycref{ax:I.VI}. +$\therefore \polygonABED = \mbox{ дважды } \polygonCED$ \byref{ax:I.VI}. \end{center} \qed @@ -3249,26 +3248,26 @@ \chapter*{Предложения} \drawPolygon[bottom][polygonCDG]{DEF,CFE,ECG} и~с данным углом \drawAngle{I}.} \begin{center} -Сделаем $\drawUnitLine{DE} = \drawUnitLine{EG}$ \bycref{prop:I.X}. +Сделаем $\drawUnitLine{DE} = \drawUnitLine{EG}$ \byref{prop:I.X}. Проведем \drawUnitLine{CE}. -Сделаем $\drawAngle{E} = \drawAngle{I}$ \bycref{prop:I.XXIII}. +Сделаем $\drawAngle{E} = \drawAngle{I}$ \byref{prop:I.XXIII}. Проведем $\left\{ \begin{aligned} \drawUnitLine{AD} &\parallel \drawUnitLine{BE}\\ \drawUnitLine{AC} &\parallel \drawUnitLine{DE}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}. +\right\}$ \byref{prop:I.XXXI}. $\drawPolygon[bottom][polygonABED]{ABFD,DEF} = \mbox{ дважды } -\drawPolygon[bottom][polygonCED]{DEF,CFE}$ \bycref{prop:I.XLI},\\ +\drawPolygon[bottom][polygonCED]{DEF,CFE}$ \byref{prop:I.XLI},\\ но $\polygonCED = -\drawPolygon[bottom][polygonDCG]{ECG}$ \bycref{prop:I.XXXVIII}. +\drawPolygon[bottom][polygonDCG]{ECG}$ \byref{prop:I.XXXVIII}. -$\therefore \polygonABED = \polygonCDG$ \bycref{ax:I.VI}. +$\therefore \polygonABED = \polygonCDG$ \byref{ax:I.VI}. \end{center} \qed @@ -3303,10 +3302,10 @@ \chapter*{Предложения} \problem{Д}{ополнения}{\drawPolygon[bottom][polygonHBGE]{HBGE} и~\drawPolygon[bottom][polygonFCIE]{FCIE} параллелограммов на одной диагонали параллелограмма равны между собой.} \begin{center} -$\drawPolygon[bottom][polygonADC]{AEF,FCIE,IDE} = \drawPolygon[bottom][polygonABD]{AEH,HBGE,GDE}$ \bycref{prop:I.XXXIV}\\ -и $\drawPolygon[bottom][polygonAEFpIDE]{AEF,IDE} = \drawPolygon[bottom][polygonAEHpGDE]{AEH,GDE}$ \bycref{prop:I.XXXIV}. +$\drawPolygon[bottom][polygonADC]{AEF,FCIE,IDE} = \drawPolygon[bottom][polygonABD]{AEH,HBGE,GDE}$ \byref{prop:I.XXXIV}\\ +и $\drawPolygon[bottom][polygonAEFpIDE]{AEF,IDE} = \drawPolygon[bottom][polygonAEHpGDE]{AEH,GDE}$ \byref{prop:I.XXXIV}. -$\therefore \polygonFCIE = \polygonHBGE$ \bycref{ax:I.III}. +$\therefore \polygonFCIE = \polygonHBGE$ \byref{ax:I.III}. \end{center} \qed @@ -3368,19 +3367,19 @@ \chapter*{Предложения} \problem[3]{К}{ данной}{прямой линии \drawUnitLine{EG} приложить параллелограмм, равный данному треугольнику \drawPolygon[middle][polygonJKL]{JKL}, и~с углом, равным данному прямолинейному углу \drawAngle{N}.} \begin{center} -Сделаем $\drawPolygon[middle][polygonAHEF]{AHEF} = \polygonJKL$ с~$\drawAngle{F} = \drawAngle{N}$ \bycref{prop:I.XLII}\\ +Сделаем $\drawPolygon[middle][polygonAHEF]{AHEF} = \polygonJKL$ с~$\drawAngle{F} = \drawAngle{N}$ \byref{prop:I.XLII}\\ и со стороной \drawUnitLine{FE}, смежной и~являющейся продолжением \drawUnitLine{EG}. Продлим \drawUnitLine{AH} до $\drawUnitLine{BG} \parallel \drawUnitLine{HE}$\\ проведем \drawUnitLine{BE}, продлим ее до продолжения \drawUnitLine{AF};\\ проведем $\drawUnitLine{CD} \parallel \drawUnitLine{FE,EG}$ до продолжения \drawUnitLine{BG} и~продлим \drawUnitLine{HE}. -$\polygonAHEF = \drawPolygon[middle][polygonEGDI]{EGDI}$ \bycref{prop:I.XLIII},\\ -но $\polygonAHEF = \polygonJKL$ \bycref{\constref}. +$\polygonAHEF = \drawPolygon[middle][polygonEGDI]{EGDI}$ \byref{prop:I.XLIII},\\ +но $\polygonAHEF = \polygonJKL$ \byref{\constref}. $\therefore \polygonEGDI = \polygonJKL$. -И $\drawAngle{F} = \drawAngle{E} =\drawAngle{I} = \drawAngle{N}$ \bycref{prop:I.XXIX,\constref}. +И $\drawAngle{F} = \drawAngle{E} =\drawAngle{I} = \drawAngle{N}$ \byref{prop:I.XXIX,\constref}. \end{center} \qed @@ -3451,13 +3450,13 @@ \chapter*{Предложения} Проведем \drawUnitLine{AD} и~\drawUnitLine{AC}, делящие прямолинейную фигуру на треугольники. Построим $\drawPolygon{FGIH} = \drawPolygon{ADE}$\\ -с $\drawAngle{I} = \drawAngle{O}$ \bycref{prop:I.XLII}.\\ +с $\drawAngle{I} = \drawAngle{O}$ \byref{prop:I.XLII}.\\ К \drawUnitLine{GI} приложим $\drawPolygon{GJKI} = \drawPolygon{ACD}$\\ -с $\drawAngle{K} = \drawAngle{O}$ \bycref{prop:I.XLIV}.\\ +с $\drawAngle{K} = \drawAngle{O}$ \byref{prop:I.XLIV}.\\ К \drawUnitLine{JK} приложим $\drawPolygon{JLMK} = \drawPolygon{ABC}$\\ -с $\drawAngle{M} = \drawAngle{O}$ \bycref{prop:I.XLIV}.\\ +с $\drawAngle{M} = \drawAngle{O}$ \byref{prop:I.XLIV}.\\ $\therefore \drawPolygon[middle][polygonFLMH]{FGIH,GJKI,JLMK} = \polygonABCDE$\\ -и \polygonFLMH является параллелограммом \bycref{prop:I.XXIX,prop:I.XIV,prop:I.XXX}\\ +и \polygonFLMH является параллелограммом \byref{prop:I.XXIX,prop:I.XIV,prop:I.XXX}\\ с $\drawAngle{M} = \drawAngle{O}$. \end{center} @@ -3488,7 +3487,7 @@ \chapter*{Предложения} \problem{Н}{а}{данной прямой \drawUnitLine{DC} построить квадрат.} \begin{center} -Проведем $\drawUnitLine{CA} \perp \mbox{ и~} = \drawUnitLine{DC}$ \bycref{prop:I.XI,prop:I.III}. +Проведем $\drawUnitLine{CA} \perp \mbox{ и~} = \drawUnitLine{DC}$ \byref{prop:I.XI,prop:I.III}. Проведем $\drawUnitLine{AB} \parallel \drawUnitLine{DC}$ и~касающуюся \drawUnitLine{BD}, проведенную $\parallel \drawUnitLine{CA}$. @@ -3500,12 +3499,12 @@ \chapter*{Предложения} draw byLabelsOnPolygon(A, B, D, C)(ALL_LABELS, 0); stopTempAngleScale; } -$\drawUnitLine{CA} = \drawUnitLine{DC}$ \bycref{\constref},\\ -$\drawAngle{C} = \drawRightAngle$ \bycref{\constref}. +$\drawUnitLine{CA} = \drawUnitLine{DC}$ \byref{\constref},\\ +$\drawAngle{C} = \drawRightAngle$ \byref{\constref}. -$\therefore \drawAngle{D} = \drawAngle{C} = \mbox{прямому углу}$ \bycref{prop:I.XXIX}, и~оставшиеся стороны и~углы должны быть равны \bycref{prop:I.XXXIV}. +$\therefore \drawAngle{D} = \drawAngle{C} = \mbox{прямому углу}$ \byref{prop:I.XXIX}, и~оставшиеся стороны и~углы должны быть равны \byref{prop:I.XXXIV}. -И $\therefore \polygonABDC$ является квадратом \bycref{def:I.XXX}. +И $\therefore \polygonABDC$ является квадратом \byref{def:I.XXX}. \end{center} \qed @@ -3581,9 +3580,9 @@ \chapter*{Предложения} \problem{В}{ прямоугольном}{треугольнике \drawLine[bottom][triangleABC]{CA,BC,AB} квадрат гипотенузы \drawUnitLine{BC} равен сумме квадратов катетов \drawUnitLine{CA} и~\drawUnitLine{AB}.} \begin{center} -На \drawUnitLine{BC}, \drawUnitLine{CA}, \drawUnitLine{AB} построим квадраты \bycref{prop:I.XLVI}. +На \drawUnitLine{BC}, \drawUnitLine{CA}, \drawUnitLine{AB} построим квадраты \byref{prop:I.XLVI}. -Проведем $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \bycref{prop:I.XXXI},\\ +Проведем $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \byref{prop:I.XXXI},\\ также проведем \drawUnitLine{BF} и~\drawUnitLine{AI}. $\drawAngle{BCI} = \drawAngle{FCA}$. @@ -3611,18 +3610,18 @@ \chapter*{Предложения} draw byLabelsOnPolygon(B,F,C)(OMIT_LABELS_AT_STRAIGHT_ANGLES, -1); stopTempAngleScale; } -$ \bycref{prop:I.IV}. +$ \byref{prop:I.IV}. -Теперь, поскольку $\drawUnitLine{AB} \parallel \drawUnitLine{CF}$ \bycref{prop:I.XIV},\\ -$\drawPolygon[middle][polygonACFG]{LAGF,CLF} = \mbox{ дважды } \polygonBLC$ \bycref{prop:I.XLI}.\\ -И поскольку $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \bycref{\constref},\\ -$\drawPolygon[middle][polygonJMCK]{JMIK,MCI} = \mbox{ дважды } \polygonAFC$ \bycref{prop:I.XLI}. +Теперь, поскольку $\drawUnitLine{AB} \parallel \drawUnitLine{CF}$ \byref{prop:I.XIV},\\ +$\drawPolygon[middle][polygonACFG]{LAGF,CLF} = \mbox{ дважды } \polygonBLC$ \byref{prop:I.XLI}.\\ +И поскольку $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \byref{\constref},\\ +$\drawPolygon[middle][polygonJMCK]{JMIK,MCI} = \mbox{ дважды } \polygonAFC$ \byref{prop:I.XLI}. -$\therefore \polygonACFG = \polygonJMCK$ \bycref{ax:I.VI}. +$\therefore \polygonACFG = \polygonJMCK$ \byref{ax:I.VI}. Так же можно показать, что $\drawPolygon[middle][polygonABED]{ABED} = \drawPolygon[middle][polygonBJKH]{BJKH}$. -А значит, $\drawPolygon[middle][polygonABEDpACFG]{ABED,LAGF,CLF} = \drawPolygon[middle][polygonBCIH]{JMIK,MCI,BJKH}$ \bycref{ax:I.II}. +А значит, $\drawPolygon[middle][polygonABEDpACFG]{ABED,LAGF,CLF} = \drawPolygon[middle][polygonBCIH]{JMIK,MCI,BJKH}$ \byref{ax:I.II}. \end{center} @@ -3652,22 +3651,22 @@ \chapter*{Предложения} \problem[3]{Е}{сли}{в треугольнике квадрат одной стороны \drawUnitLine{BC} равен сумме квадратов двух других сторон \drawUnitLine{AB} и~\drawUnitLine{AC}, то угол \drawAngle{BAC}, заключенный между этими двумя сторонами, прямой.} \begin{center} -Проведем $\drawUnitLine{AD} \perp \drawUnitLine{AB}$ и~$= \drawUnitLine{AC}$ \bycref{prop:I.XI,prop:I.III},\\ +Проведем $\drawUnitLine{AD} \perp \drawUnitLine{AB}$ и~$= \drawUnitLine{AC}$ \byref{prop:I.XI,prop:I.III},\\ также проведем \drawUnitLine{BD}. -Поскольку $\drawUnitLine{AD} = \drawUnitLine{AC}$ \bycref{\constref},\\ +Поскольку $\drawUnitLine{AD} = \drawUnitLine{AC}$ \byref{\constref},\\ $\drawUnitLine{AD}^2 = \drawUnitLine{AC}^2$. $\therefore \drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{AC}^2 + \drawUnitLine{AB}^2$. -Но $\drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BD}^2$ \bycref{prop:I.XLVII},\\ -и $\drawUnitLine{AC}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BC}^2$ \bycref{\hypref}. +Но $\drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BD}^2$ \byref{prop:I.XLVII},\\ +и $\drawUnitLine{AC}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BC}^2$ \byref{\hypref}. $\therefore \drawUnitLine{BD}^2 = \drawUnitLine{BC}^2$. $\therefore \drawUnitLine{BD} = \drawUnitLine{BC}$. -И $\therefore \drawAngle{DAB} = \drawAngle{BAC}$ \bycref{prop:I.VIII}. +И $\therefore \drawAngle{DAB} = \drawAngle{BAC}$ \byref{prop:I.VIII}. Следовательно, \drawAngle{BAC} — прямой угол. \end{center} @@ -3782,7 +3781,7 @@ \part{Книга II} \right.$} \begin{center} -Проведем $\drawProportionalLine{GB} \perp \drawProportionalLine{GK,KL,LH} \mbox{ и~} = \drawProportionalLine{A}$ \bycref{prop:I.XI,prop:I.III}. +Проведем $\drawProportionalLine{GB} \perp \drawProportionalLine{GK,KL,LH} \mbox{ и~} = \drawProportionalLine{A}$ \byref{prop:I.XI,prop:I.III}. Достроим параллелограммы, то есть\\ проведем $\left\{ @@ -3793,7 +3792,7 @@ \part{Книга II} \nointerlineskip\hbox{\drawProportionalLine{LE}} \nointerlineskip\hbox{\drawProportionalLine{HC}}} &\parallel \drawProportionalLine{GB}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}. +\right\}$ \byref{prop:I.XXXI}. $\drawPolygon{GKDB,KLED,LHCE} = \drawPolygon{GKDB} + @@ -3841,9 +3840,9 @@ \part{Книга II} } \begin{center} -Опишем \drawPolygon[middle][polygonABED]{ACFD,CBEF} \bycref{prop:I.XLVI}. +Опишем \drawPolygon[middle][polygonABED]{ACFD,CBEF} \byref{prop:I.XLVI}. -Проведем \drawProportionalLine{CF}, параллельную \drawProportionalLine{AD} \bycref{prop:I.XXXI}. +Проведем \drawProportionalLine{CF}, параллельную \drawProportionalLine{AD} \byref{prop:I.XXXI}. $\polygonABED = \drawProportionalLine{AC,CB}^2$. @@ -3886,9 +3885,9 @@ \part{Книга II} $\drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DF} = \drawProportionalLine{DF}^2 + \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$} \begin{center} -Опишем \drawPolygon{CBED} \bycref{prop:I.XLVI}. +Опишем \drawPolygon{CBED} \byref{prop:I.XLVI}. -Опишем \drawPolygon{ACDF} \bycref{prop:I.XXXI}. +Опишем \drawPolygon{ACDF} \byref{prop:I.XXXI}. Тогда $\drawPolygon[middle][polygonABEF]{ACDF,CBED} = \polygonCBED + \polygonACDF$, но\\ $\polygonABEF = \drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DE}$\\ @@ -3948,19 +3947,19 @@ \part{Книга II} } \begin{center} -Опишем \drawLine[middle][squareABED]{AD,BA,KB,EK,FE,DF} \bycref{prop:I.XLVI}, -проведем \drawProportionalLine{BG,GD} \bycref{post:I.I}\\ +Опишем \drawLine[middle][squareABED]{AD,BA,KB,EK,FE,DF} \byref{prop:I.XLVI}, +проведем \drawProportionalLine{BG,GD} \byref{post:I.I}\\ и $\left\{ \begin{aligned} \drawProportionalLine{FG,CG} &\parallel \drawProportionalLine{EK,KB}\\ \drawProportionalLine{HG,GK} &\parallel \drawProportionalLine{DF,FE}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}. +\right\}$ \byref{prop:I.XXXI}. -$\drawAngle{GBK} = \drawAngle{FDG}$ \bycref{prop:I.V}, $\drawAngle{GBK} = \drawAngle{FGD}$ \bycref{prop:I.XXIX},\\ +$\drawAngle{GBK} = \drawAngle{FDG}$ \byref{prop:I.V}, $\drawAngle{GBK} = \drawAngle{FGD}$ \byref{prop:I.XXIX},\\ $\therefore \drawAngle{FDG} = \drawAngle{FGD}$. -$\therefore$ согласно \bycref{prop:I.VI}, \bycref{prop:I.XXIX} и \bycref{prop:I.XXXIV},\\ +$\therefore$ согласно \byref{prop:I.VI}, \byref{prop:I.XXIX} и \byref{prop:I.XXXIV},\\ $\drawFromCurrentPicture[middle][squareFGHD]{ draw byNamedPolygon(DHG); draw byNamedLineFull(G, G, 1, 0, 0, -1)(DF); @@ -3973,7 +3972,7 @@ \part{Книга II} draw byNamedLineFull(G, G, 0, 1, 0, -1)(KB); draw byLabelsOnPolygon(G, C, B, K)(ALL_LABELS, 0); } является квадратом $= \drawProportionalLine{GK}^2$,\\ -$\drawPolygon[middle]{ACGH} = \drawPolygon[middle]{GKEF} = \drawProportionalLine{DF} \cdot \drawProportionalLine{GK}$ \bycref{prop:I.XLIII}. +$\drawPolygon[middle]{ACGH} = \drawPolygon[middle]{GKEF} = \drawProportionalLine{DF} \cdot \drawProportionalLine{GK}$ \byref{prop:I.XLIII}. Но $\drawFromCurrentPicture[middle][squareABEDf]{ draw byNamedPolygon(GBC,DHG,ACGH,GKEF); @@ -4031,23 +4030,23 @@ \part{Книга II} } \begin{center} -Опишем \drawPolygon[middle][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \bycref{prop:I.XLVI}, проведем \drawProportionalLine{EB}\\ +Опишем \drawPolygon[middle][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \byref{prop:I.XLVI}, проведем \drawProportionalLine{EB}\\ и $\left\{ \begin{aligned} \drawProportionalLine{DH,HG} & \parallel \drawProportionalLine{CL,LE} \\ \drawProportionalLine{LM,KL} & \parallel \drawProportionalLine{BD,DC,CA} \\ \drawProportionalLine{AK} & \parallel \drawProportionalLine{CL,LE} \\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}. +\right\}$ \byref{prop:I.XXXI}. -$\drawPolygon{CLKA} = \drawPolygon{DCLH,BDHM}$ \bycref{prop:I.XXXVI},$\drawPolygon{MHGF} = \drawPolygon{DCLH}$ \bycref{prop:I.XLIII}. +$\drawPolygon{CLKA} = \drawPolygon{DCLH,BDHM}$ \byref{prop:I.XXXVI},$\drawPolygon{MHGF} = \drawPolygon{DCLH}$ \byref{prop:I.XLIII}. -$\therefore \mbox{\bycref{ax:I.II} } \drawPolygon[middle]{DCLH,BDHM,MHGF} = \drawPolygon{DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA}$. +$\therefore \mbox{\byref{ax:I.II} } \drawPolygon[middle]{DCLH,BDHM,MHGF} = \drawPolygon{DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA}$. -Но $\drawPolygon{HLEG} = \drawProportionalLine{DC}^2$ \bycref{prop:II.IV}\\ -и $\squareCBFE = \drawProportionalLine{BD,DC}^2$ \bycref{\constref}. +Но $\drawPolygon{HLEG} = \drawProportionalLine{DC}^2$ \byref{prop:II.IV}\\ +и $\squareCBFE = \drawProportionalLine{BD,DC}^2$ \byref{\constref}. -$\therefore \mbox{\bycref{ax:I.II} } \squareCBFE = \drawPolygon{DCLH,CLKA,HLEG}$. +$\therefore \mbox{\byref{ax:I.II} } \squareCBFE = \drawPolygon{DCLH,CLKA,HLEG}$. $\therefore \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA} + \drawProportionalLine{DC}^2 = \drawProportionalLine{CA}^2 = \drawProportionalLine{BD,DC}^2$. \end{center} @@ -4097,7 +4096,7 @@ \part{Книга II} } \begin{center} -Опишем \drawPolygon[middle][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \bycref{prop:I.XLVI}, проведем \drawProportionalLine{EB} +Опишем \drawPolygon[middle][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \byref{prop:I.XLVI}, проведем \drawProportionalLine{EB} и $\left\{ \begin{aligned} @@ -4105,14 +4104,14 @@ \part{Книга II} \drawProportionalLine{LM,KL} & \parallel \drawProportionalLine{BD,DC,CA} \\ \drawProportionalLine{AK} & \parallel \drawProportionalLine{CL,LE} \\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}. +\right\}$ \byref{prop:I.XXXI}. -$\drawPolygon{MHGF} = \drawPolygon{DCLH} = \drawPolygon{CLKA}$ \bycref{prop:I.XXXVI,prop:I.XLIII}. +$\drawPolygon{MHGF} = \drawPolygon{DCLH} = \drawPolygon{CLKA}$ \byref{prop:I.XXXVI,prop:I.XLIII}. $\therefore \drawPolygon{BDHM,DCLH,MHGF} = \drawPolygon{BDHM,DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{BD,DC,CA}$. -Но $\drawPolygon{HLEG} = \drawProportionalLine{DC}^2$ \bycref{prop:II.IV},\\ -$\therefore \squareCBFE = \drawProportionalLine{CB}^2 = \drawPolygon{BDHM,DCLH,CLKA,HLEG}$ \bycref{\constref,ax:I.II}. +Но $\drawPolygon{HLEG} = \drawProportionalLine{DC}^2$ \byref{prop:II.IV},\\ +$\therefore \squareCBFE = \drawProportionalLine{CB}^2 = \drawPolygon{BDHM,DCLH,CLKA,HLEG}$ \byref{\constref,ax:I.II}. $\therefore \drawProportionalLine{BD,DC,CA} \cdot \drawProportionalLine{BD} + \drawProportionalLine{DC}^2 = \drawProportionalLine{BD,DC}^2$. \end{center} @@ -4155,7 +4154,7 @@ \part{Книга II} } \begin{center} -Опишем \drawPolygon[middle][squareABED]{DNGH,NEFG,HGCA,GFBC}. \bycref{prop:I.XLVI}, проведем \drawProportionalLine{BD} \bycref{post:I.I}\\ +Опишем \drawPolygon[middle][squareABED]{DNGH,NEFG,HGCA,GFBC}. \byref{prop:I.XLVI}, проведем \drawProportionalLine{BD} \byref{post:I.I}\\ и $\left\{ \begin{aligned} \drawProportionalLine{GN,GC} &\parallel \drawProportionalLine{EB} \\ @@ -4163,12 +4162,12 @@ \part{Книга II} \end{aligned} \right\}$. -$\drawPolygon{HGCA} = \drawPolygon{NEFG}$ \bycref{prop:I.XLIII}. +$\drawPolygon{HGCA} = \drawPolygon{NEFG}$ \byref{prop:I.XLIII}. -Добавим $\drawPolygon{GFBC} = \drawProportionalLine{NE}^2$ к~обоим \bycref{prop:II.IV}. +Добавим $\drawPolygon{GFBC} = \drawProportionalLine{NE}^2$ к~обоим \byref{prop:II.IV}. $\drawPolygon{HGCA,GFBC} = \drawPolygon{NEFG,GFBC} = \drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE}$,\\ -$\drawPolygon{DNGH} = \drawProportionalLine{DN}^2$ \bycref{prop:II.IV},\\ +$\drawPolygon{DNGH} = \drawProportionalLine{DN}^2$ \byref{prop:II.IV},\\ $\drawPolygon{HGCA,GFBC} + \drawPolygon{NEFG,GFBC} + \drawPolygon{DNGH} = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2 = \squareABED + \drawPolygon{GFBC}$. $\drawProportionalLine{DN,NE}^2 + \drawProportionalLine{NE}^2 = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2$. @@ -4236,7 +4235,7 @@ \part{Книга II} startAutoLabeling; draw byNamedLineSeq(0)(LF,HL,EH,EA,AD,DF); stopAutoLabeling; -} \bycref{prop:I.XLVI}. +} \byref{prop:I.XLVI}. Проведем \drawProportionalLine{DE}. @@ -4259,11 +4258,11 @@ \part{Книга II} \end{aligned} \right\} & \parallel \drawProportionalLine{EH,HL,LF}\\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}. +\right\}$ \byref{prop:I.XXXI}. -$\drawProportionalLine{EH,HL,LF}^2 = \drawProportionalLine{LF}^2 + \drawProportionalLine{EH,HL}^2 + 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{LF}$ \bycref{prop:II.IV}. +$\drawProportionalLine{EH,HL,LF}^2 = \drawProportionalLine{LF}^2 + \drawProportionalLine{EH,HL}^2 + 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{LF}$ \byref{prop:II.IV}. -Но $\drawProportionalLine{HL}^2 + \drawProportionalLine{EH,HL}^2 = 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$ \bycref{prop:II.VII}. +Но $\drawProportionalLine{HL}^2 + \drawProportionalLine{EH,HL}^2 = 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$ \byref{prop:II.VII}. $\therefore \drawProportionalLine{EH,HL,LF}^2 = 4 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$. \end{center} @@ -4318,13 +4317,13 @@ \part{Книга II} Проведем \drawProportionalLine{AE} и~\drawProportionalLine{EH,HF,FB},\\ $\drawProportionalLine{FD} \parallel \drawProportionalLine{CG,GE}$, $ \drawProportionalLine{GF} \parallel \drawProportionalLine{CD,DB}$ и~проведем \drawProportionalLine{AF}. -$\drawAngle{A} = \drawAngle{AEC}$ \bycref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \bycref{prop:I.XXXII},\\ -$\drawAngle{B} = \drawAngle{DFB}$ \bycref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \bycref{prop:I.XXXII},\\ +$\drawAngle{A} = \drawAngle{AEC}$ \byref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \byref{prop:I.XXXII},\\ +$\drawAngle{B} = \drawAngle{DFB}$ \byref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \byref{prop:I.XXXII},\\ $\therefore \drawAngle{AEC,CEB} = \drawRightAngle$. -$\drawAngle{B} = \drawAngle{CEB} = \drawAngle{EFG} = \drawAngle{DFB}$ \bycref{prop:I.V,prop:I.XXIX}. +$\drawAngle{B} = \drawAngle{CEB} = \drawAngle{EFG} = \drawAngle{DFB}$ \byref{prop:I.V,prop:I.XXIX}. -Значит, $\drawProportionalLine{FD} = \drawProportionalLine{DB}$, $\drawProportionalLine{GE} = \drawProportionalLine{GF} = \drawProportionalLine{CD}$ \bycref{prop:I.VI,prop:I.XXXIV}. +Значит, $\drawProportionalLine{FD} = \drawProportionalLine{DB}$, $\drawProportionalLine{GE} = \drawProportionalLine{GF} = \drawProportionalLine{CD}$ \byref{prop:I.VI,prop:I.XXXIV}. $\drawProportionalLine{AF}^2 = \left\{ \begin{aligned} @@ -4334,7 +4333,7 @@ \part{Книга II} &= \drawProportionalLine{AE}^2 + \drawProportionalLine{EH,HF}^2 \\ &= 2 \cdot \drawProportionalLine{AC}^2 + 2 \cdot \drawProportionalLine{CD}^2 \\ \end{aligned} - \right. \mbox{ \bycref{prop:I.XLVII}.} \\ + \right. \mbox{ \byref{prop:I.XLVII}.} \\ \end{aligned} \right.$ @@ -4393,23 +4392,23 @@ \part{Книга II} \drawProportionalLine{FD,DG} & \parallel \drawProportionalLine{CE} \\ \drawProportionalLine{EF} & \parallel \drawProportionalLine{CB,BD} \\ \end{aligned} -\right\}$ \bycref{prop:I.XXXI}\\ +\right\}$ \byref{prop:I.XXXI}\\ также проведем \drawProportionalLine{AG}. -$\drawAngle{A} = \drawAngle{CEA}$ \bycref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \bycref{prop:I.XXXII},\\ -$\drawAngle{CBE} = \drawAngle{BEC}$ \bycref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \bycref{prop:I.XXXII}. +$\drawAngle{A} = \drawAngle{CEA}$ \byref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \byref{prop:I.XXXII},\\ +$\drawAngle{CBE} = \drawAngle{BEC}$ \byref{prop:I.V} $= \frac{1}{2}\drawRightAngle$ \byref{prop:I.XXXII}. $\therefore \drawAngle{CEA,BEC} = \drawRightAngle$. -$\drawAngle{BEC} = \drawAngle{G}$, $\drawAngle{FEB} = \drawAngle{CBE}$\bycref{prop:I.XXIX}. +$\drawAngle{BEC} = \drawAngle{G}$, $\drawAngle{FEB} = \drawAngle{CBE}$\byref{prop:I.XXIX}. -$\drawAngle{DBG} = \drawAngle{CBE}$\bycref{prop:I.XV}. +$\drawAngle{DBG} = \drawAngle{CBE}$\byref{prop:I.XV}. $\therefore \drawAngle{DBG} = \drawAngle{CBE} = \drawAngle{BEC} = \drawAngle{FEB} = \drawAngle{G} = \frac{1}{2}\drawRightAngle$. -И $\drawProportionalLine{BD} = \drawProportionalLine{DG}$, $\drawProportionalLine{CB,BD} = \drawProportionalLine{EF} = \drawProportionalLine{FD,DG}$, \bycref{prop:I.VI,prop:I.XXXIV}. +И $\drawProportionalLine{BD} = \drawProportionalLine{DG}$, $\drawProportionalLine{CB,BD} = \drawProportionalLine{EF} = \drawProportionalLine{FD,DG}$, \byref{prop:I.VI,prop:I.XXXIV}. -Тогда, согласно \bycref{prop:I.XLVII},\\ +Тогда, согласно \byref{prop:I.XLVII},\\ $\drawUnitLine{AG}^2 = \left\{ \begin{aligned} & \drawProportionalLine{AC,CB,BD}^2 + \drawProportionalLine{DG}^2 \mbox{ или } \drawProportionalLine{BD}^2 \\ @@ -4461,16 +4460,16 @@ \part{Книга II} } \begin{center} -Опишем \drawPolygon{AHKC,HBDK} \bycref{prop:I.XLVI}. +Опишем \drawPolygon{AHKC,HBDK} \byref{prop:I.XLVI}. -Cделаем $\drawProportionalLine{EA} = \drawProportionalLine{CE}$ \bycref{prop:I.X},\\ +Cделаем $\drawProportionalLine{EA} = \drawProportionalLine{CE}$ \byref{prop:I.X},\\ проведем \drawProportionalLine{BE},\\ -возьмем $\drawProportionalLine{EA,AF} = \drawProportionalLine{BE}$ \bycref{prop:I.III},\\ -на \drawProportionalLine{AF} опишем \drawPolygon{AHGF} \bycref{prop:I.XLVI}. +возьмем $\drawProportionalLine{EA,AF} = \drawProportionalLine{BE}$ \byref{prop:I.III},\\ +на \drawProportionalLine{AF} опишем \drawPolygon{AHGF} \byref{prop:I.XLVI}. -Продлим \drawProportionalLine{KG} \bycref{post:I.II}. +Продлим \drawProportionalLine{KG} \byref{post:I.II}. -Тогда \bycref{prop:II.VI} $\drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} + \drawProportionalLine{EA}^2 = \drawProportionalLine{EA,AF}^2 = \drawProportionalLine{BE}^2 = \drawProportionalLine{AH,HB}^2 + \drawProportionalLine{EA}^2 \therefore \drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} = \drawProportionalLine{AH,HB}^2$,\\ +Тогда \byref{prop:II.VI} $\drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} + \drawProportionalLine{EA}^2 = \drawProportionalLine{EA,AF}^2 = \drawProportionalLine{BE}^2 = \drawProportionalLine{AH,HB}^2 + \drawProportionalLine{EA}^2 \therefore \drawProportionalLine{CE,EA,AF} \cdot \drawProportionalLine{AF} = \drawProportionalLine{AH,HB}^2$,\\ или $\drawPolygon{AHKC,AHGF} = \drawPolygon{AHKC,HBDK} \therefore \drawPolygon{AHGF} = \drawPolygon{HBDK}$ $\therefore \drawProportionalLine{AH,HB} \cdot \drawProportionalLine{HB} = \drawProportionalLine{AH}^2$. @@ -4503,15 +4502,15 @@ \part{Книга II} } \begin{center} -Согласно \bycref{prop:II.IV}\\ +Согласно \byref{prop:II.IV}\\ $\drawProportionalLine{CA,AD}^2 = \drawProportionalLine{CA}^2 + \drawProportionalLine{AD}^2 + 2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD}$. Добавим $\drawProportionalLine{DB}^2$ к~обоим. $\drawProportionalLine{CA,AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{CA}^2 + \drawProportionalLine{AD}^2 + 2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD} + \drawProportionalLine{DB}^2$. -Но $\drawProportionalLine{CA,AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{BC}^2$ \bycref{prop:I.XLVII},\\ -и $\drawProportionalLine{AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{BA}^2$ \bycref{prop:I.XLVII}. +Но $\drawProportionalLine{CA,AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{BC}^2$ \byref{prop:I.XLVII},\\ +и $\drawProportionalLine{AD}^2 + \drawProportionalLine{DB}^2 = \drawProportionalLine{BA}^2$ \byref{prop:I.XLVII}. $\therefore \drawProportionalLine{BC}^2 = 2 \cdot \drawProportionalLine{CA} \cdot \drawProportionalLine{AD} + \drawProportionalLine{CA}^2 + \drawProportionalLine{BA}^2$. @@ -4562,23 +4561,23 @@ \part{Книга II} \begin{center} Предположим, перпендикуляр падает внутри треугольника,\\ -тогда \bycref{prop:II.VII}\\ +тогда \byref{prop:II.VII}\\ $\drawSizedLine{BD,DC}^2 + \drawSizedLine{BD}^2 = 2 \cdot \drawSizedLine{BD,DC} \cdot \drawSizedLine{BD} + \drawSizedLine{DC}^2$. К каждой добавим $\drawSizedLine{AD}^2$,\\ тогда $\drawSizedLine{BD,DC}^2 + \drawSizedLine{BD}^2 + \drawSizedLine{AD}^2 = 2 \cdot \drawSizedLine{BD,DC} \cdot \drawSizedLine{BD} + \drawSizedLine{DC}^2 + \drawSizedLine{AD}^2$. -$\therefore$ \bycref{prop:I.XLVII} +$\therefore$ \byref{prop:I.XLVII} $\drawSizedLine{BD,DC}^2 + \drawSizedLine{AB}^2 = 2 \cdot \drawSizedLine{BD,DC} \cdot \drawSizedLine{BD} + \drawSizedLine{CA}^2$,\\ и $\therefore \drawSizedLine{CA}^2 < \drawSizedLine{BD,DC}^2 + \drawSizedLine{AB}^2$ на $2 \cdot \drawSizedLine{BD,DC} \cdot \drawSizedLine{BD}$. -Теперь предположим, что перпендикуляр падает вовне треугольника, тогда \bycref{prop:II.VII}\\ +Теперь предположим, что перпендикуляр падает вовне треугольника, тогда \byref{prop:II.VII}\\ $\drawSizedLine{FG,GH}^2 + \drawSizedLine{FG}^2 = 2 \cdot \drawSizedLine{FG,GH} \cdot \drawSizedLine{FG} + \drawSizedLine{GH}^2$. К каждой добавим $\drawSizedLine{HE}^2$, тогда\\ $\drawSizedLine{FG,GH}^2 + \drawSizedLine{FG}^2 + \drawSizedLine{HE}^2= 2 \cdot \drawSizedLine{FG,GH} \cdot \drawSizedLine{FG} + \drawSizedLine{GH}^2 + \drawSizedLine{HE}^2$. -$\therefore$ \bycref{prop:I.XLVII} +$\therefore$ \byref{prop:I.XLVII} $\drawSizedLine{EF} + \drawSizedLine{FG}^2 = 2 \cdot \drawSizedLine{FG,GH} \cdot \drawSizedLine{FG}^2 + \drawSizedLine{EG}^2$. $\therefore \drawSizedLine{EG}^2 < \drawSizedLine{EF}^2 + \drawSizedLine{FG}^2$ на $2 \cdot \drawSizedLine{FG,GH} \cdot \drawSizedLine{FG}$. @@ -4642,11 +4641,11 @@ \part{Книга II} } \begin{center} -Сделаем $\drawPolygon{BCDE} = \drawPolygon{abcdef}$ \bycref{prop:I.XLV}. +Сделаем $\drawPolygon{BCDE} = \drawPolygon{abcdef}$ \byref{prop:I.XLV}. Продлим \drawSizedLine{EG,GB} до $\drawSizedLine{FE} = \drawSizedLine{ED}$. -Возьмем $\drawSizedLine{FE,EG} = \drawSizedLine{GB}$ \bycref{prop:I.X}. +Возьмем $\drawSizedLine{FE,EG} = \drawSizedLine{GB}$ \byref{prop:I.X}. Опишем \drawFromCurrentPicture{ @@ -4657,11 +4656,11 @@ \part{Книга II} stopAutoLabeling; stopTempScale; } -\bycref{post:I.III}\\ +\byref{post:I.III}\\ и продлим до нее \drawSizedLine{ED}: проведем \drawSizedLine{HG}. -$\drawSizedLine{BG}^2 \mbox{ или } \drawSizedLine{HG}^2 = \drawSizedLine{FE} \cdot \drawSizedLine{EG,GB} + \drawSizedLine{EG}^2$ \bycref{prop:II.V},\\ -но $\drawSizedLine{HG}^2 = \drawSizedLine{HE} ^2 + \drawSizedLine{EG}^2$ \bycref{prop:I.XLVII}. +$\drawSizedLine{BG}^2 \mbox{ или } \drawSizedLine{HG}^2 = \drawSizedLine{FE} \cdot \drawSizedLine{EG,GB} + \drawSizedLine{EG}^2$ \byref{prop:II.V},\\ +но $\drawSizedLine{HG}^2 = \drawSizedLine{HE} ^2 + \drawSizedLine{EG}^2$ \byref{prop:I.XLVII}. $\therefore \drawSizedLine{HE}^2 + \drawSizedLine{EG}^2 = \drawSizedLine{FE} \cdot \drawSizedLine{EG,GB} + \drawSizedLine{EG}^2$. @@ -4911,11 +4910,11 @@ \chapter*{Определения} Действительно, пусть это не так, тогда пусть другая точка будет центром, и~проведем от нее \drawUnitLine{GA}, \drawUnitLine{GD} и~\drawUnitLine{GB}. -Поскольку в~\drawLine{AG,DG,AD} и~\drawLine{DB,DG,BG} $\drawUnitLine{AG} = \drawUnitLine{BG}$ \bycref{\hypref,def:I.XV},\\ -$\drawUnitLine{AD} = \drawUnitLine{DB}$ \bycref{\constref} и~\drawUnitLine{DG} общая обоим,\\ -$\drawAngle{ADF,FDG} = \drawAngle{GDB}$ \bycref{prop:I.VIII} и, следовательно, являются прямыми углами. +Поскольку в~\drawLine{AG,DG,AD} и~\drawLine{DB,DG,BG} $\drawUnitLine{AG} = \drawUnitLine{BG}$ \byref{\hypref,def:I.XV},\\ +$\drawUnitLine{AD} = \drawUnitLine{DB}$ \byref{\constref} и~\drawUnitLine{DG} общая обоим,\\ +$\drawAngle{ADF,FDG} = \drawAngle{GDB}$ \byref{prop:I.VIII} и, следовательно, являются прямыми углами. -Но $\drawAngle{FDG,GDB} = \drawRightAngle$ \bycref{\constref}, $\drawAngle{GDB} = \drawAngle{FDG,GDB}$ \bycref{ax:I.XI},\\ +Но $\drawAngle{FDG,GDB} = \drawRightAngle$ \byref{\constref}, $\drawAngle{GDB} = \drawAngle{FDG,GDB}$ \byref{ax:I.XI},\\ что невозможно. \end{center} @@ -4951,16 +4950,16 @@ \chapter*{Определения} \problem[2]{П}{рямая}{\drawSizedLine{AB}, соединяющая две точки на окружности круга \drawCircle[middle][1/4]{D}, целиком находится внутри круга.} \begin{center} -Найдем центр \circleD\ \bycref{prop:III.I}. +Найдем центр \circleD\ \byref{prop:III.I}. Из центра проведем \drawSizedLine{DE} к~любой точке \drawSizedLine{AB}, пересекающую окружность. Проведем \drawSizedLine{AD} и~\drawSizedLine{BD}. -Тогда $\drawAngle{A} = \drawAngle{B}$ \bycref{prop:I.V},\\ -но $\drawAngle{E} > \drawAngle{A} \mbox{ или } > \drawAngle{B}$ \bycref{prop:I.XVI}. +Тогда $\drawAngle{A} = \drawAngle{B}$ \byref{prop:I.V},\\ +но $\drawAngle{E} > \drawAngle{A} \mbox{ или } > \drawAngle{B}$ \byref{prop:I.XVI}. -$\therefore \drawSizedLine{AD} > \drawSizedLine{DE}$ \bycref{prop:I.XIX},\\ +$\therefore \drawSizedLine{AD} > \drawSizedLine{DE}$ \byref{prop:I.XIX},\\ но $\drawSizedLine{AD} = \drawSizedLine{DE,EF}$. $\therefore \drawSizedLine{DE,EF} > \drawSizedLine{DE}$. @@ -5012,17 +5011,17 @@ \chapter*{Определения} $\drawUnitLine{AE} = \drawUnitLine{EB}$, \drawUnitLine{EF} общая\\ и $\drawUnitLine{AF} = \drawUnitLine{FB}$. -$\therefore \drawAngle{AFE} = \drawAngle{EFB}$ \bycref{prop:I.VIII}\\ -и $\therefore \drawUnitLine{EF} \perp \drawUnitLine{AF,FB}$ \bycref{def:I.X}. +$\therefore \drawAngle{AFE} = \drawAngle{EFB}$ \byref{prop:I.VIII}\\ +и $\therefore \drawUnitLine{EF} \perp \drawUnitLine{AF,FB}$ \byref{def:I.X}. Теперь пусть $\drawUnitLine{EF} \perp \drawUnitLine{AF,FB}$. Тогда в~ \triangleAEF\ и~\triangleFEB\\ -$\drawAngle{A} = \drawAngle{B}$ \bycref{prop:I.V},\\ -$\drawAngle{AFE} = \drawAngle{EFB}$ \bycref{\hypref}\\ +$\drawAngle{A} = \drawAngle{B}$ \byref{prop:I.V},\\ +$\drawAngle{AFE} = \drawAngle{EFB}$ \byref{\hypref}\\ и $\drawUnitLine{AE} = \drawUnitLine{EB}$. -$\therefore \drawUnitLine{AF} = \drawUnitLine{FB}$ \bycref{prop:I.XXVI}. +$\therefore \drawUnitLine{AF} = \drawUnitLine{FB}$ \byref{prop:I.XXVI}. И $\therefore \drawUnitLine{EF}$ рассекает \drawUnitLine{AF,FB} пополам. \end{center} @@ -5059,11 +5058,11 @@ \chapter*{Определения} Но если ни одна из прямых \drawUnitLine{AC} или \drawUnitLine{BD} не проходит через центр, проведем \drawUnitLine{EF} из центра к~точке их пересечения. \begin{center} -Если \drawUnitLine{AC} делится пополам, \drawUnitLine{EF} $\perp$ ей \bycref{prop:III.III}. +Если \drawUnitLine{AC} делится пополам, \drawUnitLine{EF} $\perp$ ей \byref{prop:III.III}. $\therefore \drawAngle{FED,DEC} = \drawRightAngle$. -И если \drawUnitLine{BD} делится пополам, $\drawUnitLine{EF} \perp \drawUnitLine{BD}$ \bycref{prop:III.III}. +И если \drawUnitLine{BD} делится пополам, $\drawUnitLine{EF} \perp \drawUnitLine{BD}$ \byref{prop:III.III}. $\therefore \drawAngle{FED} = \drawRightAngle$. @@ -5110,8 +5109,8 @@ \chapter*{Определения} Допустим, это возможно, и~два пересекающихся круга имеют общий центр. Из предполагаемого центра проведем \drawUnitLine{CE} к~точке пересечения и~\drawUnitLine{EF,FG}, пересекающую окружности. \begin{center} -Тогда $\drawUnitLine{CE} = \drawUnitLine{EF}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{CE} = \drawUnitLine{EF,FG}$ \bycref{def:I.XV}. +Тогда $\drawUnitLine{CE} = \drawUnitLine{EF}$ \byref{def:I.XV}\\ +и $\drawUnitLine{CE} = \drawUnitLine{EF,FG}$ \byref{def:I.XV}. $\therefore \drawUnitLine{EF} = \drawUnitLine{EF,FG}$\\ часть равна целому, что невозможно. @@ -5154,8 +5153,8 @@ \chapter*{Определения} Действительно, пусть это возможно, и~у кругов будет один центр. Из предполагаемого центра проведем \drawUnitLine{FE,EB} и~\drawUnitLine{CF} к~точке касания. \begin{center} -Тогда $\drawUnitLine{CF} = \drawUnitLine{FE}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{CF} = \drawUnitLine{FE,EB}$ \bycref{def:I.XV}. +Тогда $\drawUnitLine{CF} = \drawUnitLine{FE}$ \byref{def:I.XV}\\ +и $\drawUnitLine{CF} = \drawUnitLine{FE,EB}$ \byref{def:I.XV}. $\therefore \drawUnitLine{FE} = \drawUnitLine{FE,EB}$. \end{center} @@ -5234,11 +5233,11 @@ \chapter*{Определения} } \drawCurrentPictureInMargin \startsubproposition{Часть I} -Из центра круга проведем \drawUnitLine{EB} и~\drawUnitLine{EC}, тогда $\drawUnitLine{AE} = \drawUnitLine{EB}$ \bycref{def:I.XV}, $\drawUnitLine{EF,AE} = \drawUnitLine{EF} + \drawUnitLine{EB} > \drawUnitLine{FB}$ \bycref{prop:I.XX}. Таким же образом можно показать, что \drawUnitLine{EF,AE} больше \drawUnitLine{FC} или любой другой линии, проведенной из той же точки к~окружности. +Из центра круга проведем \drawUnitLine{EB} и~\drawUnitLine{EC}, тогда $\drawUnitLine{AE} = \drawUnitLine{EB}$ \byref{def:I.XV}, $\drawUnitLine{EF,AE} = \drawUnitLine{EF} + \drawUnitLine{EB} > \drawUnitLine{FB}$ \byref{prop:I.XX}. Таким же образом можно показать, что \drawUnitLine{EF,AE} больше \drawUnitLine{FC} или любой другой линии, проведенной из той же точки к~окружности. -Теперь согласно \bycref{prop:I.XX} $\drawUnitLine{EF} + \drawUnitLine{FC} > \drawUnitLine{EC} = \drawUnitLine{FD} + \drawUnitLine{EF}$, вычтем \drawUnitLine{EF} из обеих. $\therefore \drawUnitLine{FC} > \drawUnitLine{FD}$ \bycref{ax:I.III}, и~так же можно показать, что \drawUnitLine{FD} меньше любой другой линии, проведенной из той же точки к~окружности. +Теперь согласно \byref{prop:I.XX} $\drawUnitLine{EF} + \drawUnitLine{FC} > \drawUnitLine{EC} = \drawUnitLine{FD} + \drawUnitLine{EF}$, вычтем \drawUnitLine{EF} из обеих. $\therefore \drawUnitLine{FC} > \drawUnitLine{FD}$ \byref{ax:I.III}, и~так же можно показать, что \drawUnitLine{FD} меньше любой другой линии, проведенной из той же точки к~окружности. -Теперь в~\drawLine[middle][triangleEFB]{FB,EF,EB} и~\drawLine[middle][triangleEFC]{FC,EF,EC} \drawUnitLine{EF} общая, $\drawAngle{BEC,CEF} > \drawAngle{CEF}$ и~$\drawUnitLine{EB} = \drawUnitLine{EC} \therefore \drawUnitLine{FB} > \drawUnitLine{FC}$ \bycref{prop:I.XXIV} и~так же можно показать, что \drawUnitLine{FB} больше любой другой линии, проведенной из той же точки к~окружности и~проходящей дальше \drawUnitLine{EF,AE}. +Теперь в~\drawLine[middle][triangleEFB]{FB,EF,EB} и~\drawLine[middle][triangleEFC]{FC,EF,EC} \drawUnitLine{EF} общая, $\drawAngle{BEC,CEF} > \drawAngle{CEF}$ и~$\drawUnitLine{EB} = \drawUnitLine{EC} \therefore \drawUnitLine{FB} > \drawUnitLine{FC}$ \byref{prop:I.XXIV} и~так же можно показать, что \drawUnitLine{FB} больше любой другой линии, проведенной из той же точки к~окружности и~проходящей дальше \drawUnitLine{EF,AE}. \defineNewPicture[1/2]{ pair A, B, C, D, E, F, G, H, M; @@ -5280,7 +5279,7 @@ \chapter*{Определения} Тогда в~\drawLine[middle][triangleEFM]{EM,EF,FM} и~\drawLine[middle][triangleEFB]{FB,EF,EB} \drawUnitLine{EF} общая,\\ $\drawAngle{GFE} = \drawAngle{BFE}$ и~$\drawUnitLine{FB} = \drawUnitLine{FM}$. -$\therefore \drawUnitLine{EB} = \drawUnitLine{EM}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{EB} = \drawUnitLine{EM}$ \byref{prop:I.IV}. $\therefore \drawUnitLine{EB} = \drawUnitLine{EH} = \drawUnitLine{EM}$,\\ часть равна целому, что невозможно. @@ -5325,11 +5324,11 @@ \chapter*{Определения} Тогда \drawUnitLine{DM,MA}, проходящая через центр, будет наибольшей,\\ поскольку, раз $\drawUnitLine{MA} = \drawUnitLine{ME}$, если к~обеим добавить \drawUnitLine{DM}, $\drawUnitLine{DM,MA} = \drawUnitLine{DM} + \drawUnitLine{ME}$, -но $> \drawUnitLine{DE}$ \bycref{prop:I.XX},\\ +но $> \drawUnitLine{DE}$ \byref{prop:I.XX},\\ $\therefore$ \drawUnitLine{DM,MA} больше любой другой линии, проведенной из той же точки к~вогнутой части окружности. Теперь в~\drawLine{DM,MF,DF} и~\drawLine{DM,ME,DE} $\drawUnitLine{MF} = \drawUnitLine{ME}$ и~\drawUnitLine{DM} общая,\\ -но $\drawAngle{DMF,FME} > \drawAngle{DMF}$, $\therefore \drawUnitLine{DE} > \drawUnitLine{DF}$ \bycref{prop:I.XXIV}. +но $\drawAngle{DMF,FME} > \drawAngle{DMF}$, $\therefore \drawUnitLine{DE} > \drawUnitLine{DF}$ \byref{prop:I.XXIV}. Так же можно показать, что $\drawUnitLine{DE} >$ любой другой линии, более далекой от \drawUnitLine{DM,MA}. \end{center} @@ -5364,11 +5363,11 @@ \chapter*{Определения} Из линий, падающих на выпуклую часть окружности, наименьшей \drawUnitLine{DG} является та, которая при продолжении проходит через центр, а~линия, ближняя к~наименьшей, будет меньше дальней. \begin{center} -Действительно, поскольку $\drawUnitLine{KM} + \drawUnitLine{DK} > \drawUnitLine{DG,GM}$ \bycref{prop:I.XX}\\ +Действительно, поскольку $\drawUnitLine{KM} + \drawUnitLine{DK} > \drawUnitLine{DG,GM}$ \byref{prop:I.XX}\\ и $\drawUnitLine{KM} = \drawUnitLine{GM}$,\\ -$\therefore \drawUnitLine{DK} > \drawUnitLine{DG}$ \bycref{ax:I.V}. +$\therefore \drawUnitLine{DK} > \drawUnitLine{DG}$ \byref{ax:I.V}. -Значит, поскольку $\drawUnitLine{HM} + \drawUnitLine{DH} > \drawUnitLine{KM} + \drawUnitLine{DK}$ \bycref{prop:I.XXI},\\ +Значит, поскольку $\drawUnitLine{HM} + \drawUnitLine{DH} > \drawUnitLine{KM} + \drawUnitLine{DK}$ \byref{prop:I.XXI},\\ и $\drawUnitLine{HM} = \drawUnitLine{KM}$,\\ $\therefore \drawUnitLine{DK} < \drawUnitLine{DH}$. Так же и~для прочих. \end{center} @@ -5416,7 +5415,7 @@ \chapter*{Определения} Тогда в~\drawLine{DM,DO,BO,BM} и~\drawLine{HM,DH,DM} получим $\drawUnitLine{DO} = \drawUnitLine{DH}$\\ и общую \drawUnitLine{DM}, а~также $\drawAngle{MDN} = \drawAngle{HDM}$. -$\therefore \drawUnitLine{BM,BO} = \drawUnitLine{HM}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{BM,BO} = \drawUnitLine{HM}$ \byref{prop:I.IV}. Но $\drawUnitLine{HM} = \drawUnitLine{BM}$. @@ -5462,7 +5461,7 @@ \chapter*{Определения} stopGlobalRotation; } должна быть центром, проведем между этими двумя точками \drawUnitLine{DF} и~продлим в~обе стороны до окружности. -Теперь, поскольку более чем две равных прямых линии проведено к~окружности из точки, не являющейся центром, две из них должны лежать по одну сторону диаметра \drawUnitLine{DH,DF,FL}, и, поскольку из точки \drawPointL[middle][DA,DF]{D}, не являющейся центром, прямые линии проведены к~окружности, наибольшая из них \drawUnitLine{DF,FL}, та, что проходит через центр, а~\drawUnitLine{DC}, которая ближе к~\drawUnitLine{DF,FL}, $> \drawUnitLine{DB}$, расположенной дальше \bycref{prop:III.VIII}, но $\drawUnitLine{DC} = \drawUnitLine{DB}$ \bycref{\hypref}, что невозможно. +Теперь, поскольку более чем две равных прямых линии проведено к~окружности из точки, не являющейся центром, две из них должны лежать по одну сторону диаметра \drawUnitLine{DH,DF,FL}, и, поскольку из точки \drawPointL[middle][DA,DF]{D}, не являющейся центром, прямые линии проведены к~окружности, наибольшая из них \drawUnitLine{DF,FL}, та, что проходит через центр, а~\drawUnitLine{DC}, которая ближе к~\drawUnitLine{DF,FL}, $> \drawUnitLine{DB}$, расположенной дальше \byref{prop:III.VIII}, но $\drawUnitLine{DC} = \drawUnitLine{DB}$ \byref{\hypref}, что невозможно. То же можно показать для любой другой точки, кроме \drawPointL[middle][DA,DF]{D}, которая, таким образом, должна быть центром круга. @@ -5527,12 +5526,12 @@ \chapter*{Определения} Из центра \drawCircle[middle][1/5]{PII} проведем \drawUnitLine{PG}, \drawUnitLine{PB} и~\drawUnitLine{PH} к~точкам пересечения. -$\therefore \drawUnitLine{PG} = \drawUnitLine{PB} = \drawUnitLine{PH}$ \bycref{def:I.XV},\\ -но, поскольку круги пересекаются, у~них не один центр \bycref{prop:III.V}. +$\therefore \drawUnitLine{PG} = \drawUnitLine{PB} = \drawUnitLine{PH}$ \byref{def:I.XV},\\ +но, поскольку круги пересекаются, у~них не один центр \byref{prop:III.V}. $\therefore$ рассматриваемая точка не центр \circlePI,\\ и~$\therefore$, поскольку \drawUnitLine{PG}, \drawUnitLine{PB} и~\drawUnitLine{PH} проведены не из центра,\\ -они не равны \bycref{prop:III.VII,prop:III.VIII}. +они не равны \byref{prop:III.VII,prop:III.VIII}. Но выше было показано, что они равны, что невозможно, круги, стало быть, не пересекаются в~трех точках. \end{center} @@ -5588,7 +5587,7 @@ \chapter*{Определения} stopAutoLabeling; draw byNamedLineFull(A, A, 1, 1, 0, 0)(GF); } -$\drawSizedLine{GF} + \drawSizedLine{AG} > \drawSizedLine{AF}$ \bycref{prop:I.XX},\\ +$\drawSizedLine{GF} + \drawSizedLine{AG} > \drawSizedLine{AF}$ \byref{prop:I.XX},\\ а $\drawSizedLine{AF} = \drawSizedLine{HD,DG,GF}$, как радиусы\circleM. Но $\drawSizedLine{GF} + \drawSizedLine{AG} > \drawSizedLine{HD,DG,GF}$. @@ -5643,9 +5642,9 @@ \chapter*{Определения} Действительно, если такое возможно, соединим центры с~помощью \drawUnitLine{FC,CD,DG}, не проходящей через точку касания, из точки касания проведем \drawUnitLine{AF} и~\drawUnitLine{AG} к~центрам. \begin{center} -Поскольку $\drawUnitLine{AF} + \drawUnitLine{AG} > \drawUnitLine{FC,CD,DG}$ \bycref{prop:I.XX},\\ -$\drawUnitLine{FC} = \drawUnitLine{AF}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{DG} = \drawUnitLine{AG}$ \bycref{def:I.XV},\\ +Поскольку $\drawUnitLine{AF} + \drawUnitLine{AG} > \drawUnitLine{FC,CD,DG}$ \byref{prop:I.XX},\\ +$\drawUnitLine{FC} = \drawUnitLine{AF}$ \byref{def:I.XV}\\ +и $\drawUnitLine{DG} = \drawUnitLine{AG}$ \byref{def:I.XV},\\ $\therefore \drawUnitLine{FC} + \drawUnitLine{DG} > \drawUnitLine{FC,CD,DG}$, часть больше целого, что невозможно. \end{center} @@ -5690,20 +5689,20 @@ \chapter*{Определения} \startsubproposition{Случай I.} \drawCurrentPictureInMargin -Действительно, если это возможно, пусть \drawCircle{M} и~\drawCircle{N} касаются друг друга внутри в~двух точках, проведем \drawUnitLine{GH}, соединяющую их центры, и~продлим до одной из точек касания \bycref{prop:III.XI}. +Действительно, если это возможно, пусть \drawCircle{M} и~\drawCircle{N} касаются друг друга внутри в~двух точках, проведем \drawUnitLine{GH}, соединяющую их центры, и~продлим до одной из точек касания \byref{prop:III.XI}. \begin{center} Проведем \drawUnitLine{DH} и~\drawUnitLine{DG}. -Но $\drawUnitLine{BG} = \drawUnitLine{DG}$ \bycref{def:I.XV}. +Но $\drawUnitLine{BG} = \drawUnitLine{DG}$ \byref{def:I.XV}. $\therefore$ если добавить к~каждой \drawUnitLine{GH},\\ $\drawUnitLine{GH,BG} = \drawUnitLine{GH} + \drawUnitLine{DG}$. -Но $\drawUnitLine{GH,BG} = \drawUnitLine{DH}$ \bycref{def:I.XV}\\ +Но $\drawUnitLine{GH,BG} = \drawUnitLine{DH}$ \byref{def:I.XV}\\ и $\therefore \drawUnitLine{GH} + \drawUnitLine{DG} = \drawUnitLine{DH}$. -Но $\therefore \drawUnitLine{GH} + \drawUnitLine{DG} > \drawUnitLine{DH}$ \bycref{prop:I.XX},\\ +Но $\therefore \drawUnitLine{GH} + \drawUnitLine{DG} > \drawUnitLine{DH}$ \byref{prop:I.XX},\\ что невозможно. \end{center} @@ -5767,12 +5766,12 @@ \chapter*{Определения} Теперь, если такое возможно, пусть круги \drawCircle{H} и~\drawCircle{G} касаются друг друга снаружи в~двух точках. Проведем \drawUnitLine{HA,AG}, соединяющую центры и~проходящую через одну из точек касания, проведем также \drawUnitLine{CH} и~\drawUnitLine{CG}. \begin{center} -$\drawUnitLine{CH} = \drawUnitLine{HA}$ \bycref{def:I.XV}\\ -и $\drawUnitLine{AG} = \drawUnitLine{CG}$ \bycref{def:I.XV}. +$\drawUnitLine{CH} = \drawUnitLine{HA}$ \byref{def:I.XV}\\ +и $\drawUnitLine{AG} = \drawUnitLine{CG}$ \byref{def:I.XV}. $\therefore \drawUnitLine{CG} + \drawUnitLine{CH} = \drawUnitLine{HA,AG}$. -Но $\drawUnitLine{CG} + \drawUnitLine{CH} > \drawUnitLine{HA,AG}$ \bycref{prop:I.XX}, что невозможно. +Но $\drawUnitLine{CG} + \drawUnitLine{CH} > \drawUnitLine{HA,AG}$ \byref{prop:I.XX}, что невозможно. \end{center} Следовательно, ни в~каком случае круги не касаются друг друга в~двух точках. @@ -5821,16 +5820,16 @@ \chapter*{Определения} проведем $\drawProportionalLine{EF} \perp \drawProportionalLine{AF,FB}$ и~$\drawProportionalLine{EG} \perp \drawProportionalLine{CG,GD}$,\\ проведем \drawProportionalLine{EA} и~\drawProportionalLine{EC}. -Тогда $\drawProportionalLine{CG} = \frac{1}{2} \drawProportionalLine{CG,GD}$ \bycref{prop:III.III}\\ -и $\drawProportionalLine{AF} = \frac{1}{2} \drawProportionalLine{AF,FB}$ \bycref{prop:III.III}. +Тогда $\drawProportionalLine{CG} = \frac{1}{2} \drawProportionalLine{CG,GD}$ \byref{prop:III.III}\\ +и $\drawProportionalLine{AF} = \frac{1}{2} \drawProportionalLine{AF,FB}$ \byref{prop:III.III}. -Поскольку $\drawProportionalLine{CG,GD} = \drawProportionalLine{AF,FB}$ \bycref{\hypref},\\ -$\therefore \drawProportionalLine{CG} = \drawProportionalLine{AF}$ \bycref{ax:I.VII}\\ -и $\drawProportionalLine{EA} = \drawProportionalLine{EC}$ \bycref{def:I.XV},\\ +Поскольку $\drawProportionalLine{CG,GD} = \drawProportionalLine{AF,FB}$ \byref{\hypref},\\ +$\therefore \drawProportionalLine{CG} = \drawProportionalLine{AF}$ \byref{ax:I.VII}\\ +и $\drawProportionalLine{EA} = \drawProportionalLine{EC}$ \byref{def:I.XV},\\ $\therefore \drawProportionalLine{EA}^2 = \drawProportionalLine{EC}^2$. -Но поскольку \drawAngle{F} прямой \bycref{\constref},\\ -$\drawProportionalLine{EA}^2 = \drawProportionalLine{EF}^2 + \drawProportionalLine{AF}^2$ \bycref{prop:I.XLVII}\\ +Но поскольку \drawAngle{F} прямой \byref{\constref},\\ +$\drawProportionalLine{EA}^2 = \drawProportionalLine{EF}^2 + \drawProportionalLine{AF}^2$ \byref{prop:I.XLVII}\\ и $\drawProportionalLine{EC}^2 = \drawProportionalLine{EG}^2 + \drawProportionalLine{CG}^2$ по той же причине. $\therefore \drawProportionalLine{EF}^2 + \drawProportionalLine{AF}^2 = \drawProportionalLine{EG}^2 + \drawProportionalLine{CG}^2$. @@ -5893,7 +5892,7 @@ \chapter*{Определения} Тогда $\drawUnitLine{EM} = \drawUnitLine{EA}$\\ и $\drawUnitLine{EN} = \drawUnitLine{DE}$,\\ $\therefore \drawUnitLine{EN} + \drawUnitLine{EM} = \drawUnitLine{DE,EA}$,\\ -но $\drawUnitLine{EN} + \drawUnitLine{EM} > \drawUnitLine{MN}$ \bycref{prop:I.XX}. +но $\drawUnitLine{EN} + \drawUnitLine{EM} > \drawUnitLine{MN}$ \byref{prop:I.XX}. $\therefore \drawUnitLine{DE,EA} > \drawUnitLine{MN}$. \end{center} @@ -5922,7 +5921,7 @@ \chapter*{Определения} $\drawUnitLine{EN} \mbox{ и~} \drawUnitLine{EM} = \drawUnitLine{EG} \mbox{ и~} \drawUnitLine{EF}$,\\ но $\drawAngle{NEG,GEF,FEM} > \drawAngle{GEF}$. -$\therefore \drawUnitLine{MN} > \drawUnitLine{EF}$ \bycref{prop:I.XXIV}. +$\therefore \drawUnitLine{MN} > \drawUnitLine{EF}$ \byref{prop:I.XXIV}. \end{center} \vfill\pagebreak @@ -5968,7 +5967,7 @@ \chapter*{Определения} Сделаем $\drawUnitLine{EH} = \drawUnitLine{EL}$\\ и проведем $\drawUnitLine{NM} \perp \drawUnitLine{EL,LK}$. -Поскольку \drawUnitLine{BC} и~\drawUnitLine{NM} равноудалены от~центра, $\drawUnitLine{BC} = \drawUnitLine{NM}$ \bycref{prop:III.XIV},\\ +Поскольку \drawUnitLine{BC} и~\drawUnitLine{NM} равноудалены от~центра, $\drawUnitLine{BC} = \drawUnitLine{NM}$ \byref{prop:III.XIV},\\ но $\drawUnitLine{NM} > \drawUnitLine{GF}$ (случай I). $\therefore \drawUnitLine{BC} > \drawUnitLine{GF}$. @@ -6021,9 +6020,9 @@ \chapter*{Определения} \begin{center} Тогда, поскольку $\drawUnitLine{DA} = \drawUnitLine{DC}$,\\ -$\drawAngle{CAD} = \drawAngle{C}$ \bycref{prop:I.V},\\ -и $\therefore$ оба угла острые \bycref{prop:I.XVII},\\ -но $\drawAngle{CAD} = \drawRightAngle$ \bycref{\hypref}, что невозможно. +$\drawAngle{CAD} = \drawAngle{C}$ \byref{prop:I.V},\\ +и $\therefore$ оба угла острые \byref{prop:I.XVII},\\ +но $\drawAngle{CAD} = \drawRightAngle$ \byref{\hypref}, что невозможно. Следовательно, \drawUnitLine{AC}, проведенная $\perp \drawUnitLine{DA}$, не пересекает окружность в~другом месте. \end{center} @@ -6086,7 +6085,7 @@ \chapter*{Определения} \drawCurrentPictureInMargin \problem{П}{ровести}{касательную к~данному кругу \drawCircle[middle][1/5]{EI} из данной точки.} -Если данная точка \drawPointL[middle][FB]{B} расположена на окружности, ясно, что прямая $\drawUnitLine{AB} \perp \drawUnitLine{BE}$ радиусу и~будет искомой касательной \bycref{prop:III.XVI}. +Если данная точка \drawPointL[middle][FB]{B} расположена на окружности, ясно, что прямая $\drawUnitLine{AB} \perp \drawUnitLine{BE}$ радиусу и~будет искомой касательной \byref{prop:III.XVI}. \begin{center} Но если точка \drawPointL[middle][FB]{A} расположена вовне, проведем из нее \drawUnitLine{AD,DE} к~центру, секущую \circleEI,\\ @@ -6099,7 +6098,7 @@ \chapter*{Определения} Тогда \drawUnitLine{AB} и~будет искомой касательной.\\ Поскольку в~\drawLine[bottom]{FD,FB,BE,DE} и~\drawLine[bottom]{BE,DE,AD,AB} $\drawUnitLine{AD,DE} = \drawUnitLine{FB,BE}$, \drawAngle{E} общий\\ и $\drawUnitLine{DE} = \drawUnitLine{BE}$,\\ -$\therefore \mbox{ \bycref{prop:I.IV} } \drawAngle{B} = \drawAngle{D} = \drawRightAngle$,\\ +$\therefore \mbox{ \byref{prop:I.IV} } \drawAngle{B} = \drawAngle{D} = \drawRightAngle$,\\ $\therefore \drawUnitLine{FD}$ касательная к~\circleEI. \end{center} @@ -6135,9 +6134,9 @@ \chapter*{Определения} \begin{center} Действительно, если возможно, пусть \drawUnitLine{FB,BG} будет $\perp \drawUnitLine{CD}$,\\ -тогда, поскольку $\drawAngle{G} = \drawRightAngle$, \drawAngle{C} острый \bycref{prop:I.XVII}. +тогда, поскольку $\drawAngle{G} = \drawRightAngle$, \drawAngle{C} острый \byref{prop:I.XVII}. -$\therefore \drawUnitLine{FC} > \drawUnitLine{FB,BG}$ \bycref{prop:I.XIX}. +$\therefore \drawUnitLine{FC} > \drawUnitLine{FB,BG}$ \byref{prop:I.XIX}. Но $\drawUnitLine{FC} = \drawUnitLine{FB}$,\\ и $\therefore \drawUnitLine{FB} > \drawUnitLine{FB,BG}$, часть больше целого, что невозможно. @@ -6176,10 +6175,10 @@ \chapter*{Определения} Действительно, если центр не находится на \drawUnitLine{AC}, проведем \drawUnitLine{CF} к~предполагаемому центру из точки касания. \begin{center} -Поскольку $\drawUnitLine{CF} \perp \drawUnitLine{CE}$ \bycref{prop:III.XVIII},\\ +Поскольку $\drawUnitLine{CF} \perp \drawUnitLine{CE}$ \byref{prop:III.XVIII},\\ $\therefore \drawAngle{FCE} = \drawRightAngle$, прямому углу. -Но $\drawAngle{ACF,FCE} = \drawRightAngle$ \bycref{\hypref}. +Но $\drawAngle{ACF,FCE} = \drawRightAngle$ \byref{\hypref}. И $\therefore \drawAngle{FCE} = \drawAngle{ACF,FCE}$,\\ часть равна целому, что невозможно. @@ -6219,10 +6218,10 @@ \chapter*{Определения} стороне \drawAngle{CAF}. Поскольку $\drawUnitLine{EC} = \drawUnitLine{EA}$,\\ -$\drawAngle{CAF} = \drawAngle{C}$ \bycref{prop:I.V}. +$\drawAngle{CAF} = \drawAngle{C}$ \byref{prop:I.V}. Но $\drawAngle{CEF} = \drawAngle{CAF} + \drawAngle{C}$,\\ -или $\drawAngle{CEF} = \mbox{ дважды } \drawAngle{CAF}$ \bycref{prop:I.XXXII}. +или $\drawAngle{CEF} = \mbox{ дважды } \drawAngle{CAF}$ \byref{prop:I.XXXII}. \end{center} \defineNewPicture{ @@ -6256,7 +6255,7 @@ \chapter*{Определения} Проведем \drawUnitLine{AF} из угла через центр круга. -Тогда $\drawAngle{B} = \drawAngle{BAF}$ и~$\drawAngle{C} = \drawAngle{CAF}$, вследствие равенства сторон \bycref{prop:I.V}. +Тогда $\drawAngle{B} = \drawAngle{BAF}$ и~$\drawAngle{C} = \drawAngle{CAF}$, вследствие равенства сторон \byref{prop:I.V}. Значит, $\drawAngle{BAF} + \drawAngle{B} + \drawAngle{CAF} + \drawAngle{C} = \mbox{ дважды } \drawAngle{BAF,CAF}$. @@ -6334,7 +6333,7 @@ \chapter*{Определения} Пусть сегмент будет больше половины круга, проведем \drawUnitLine{DF} и~\drawUnitLine{BF} к~центру. \begin{center} -$\drawAngle{F} = \mbox{ дважды } \drawAngle{A} \mbox{ или дважды } = \drawAngle{E}$ \bycref{prop:III.XX}. +$\drawAngle{F} = \mbox{ дважды } \drawAngle{A} \mbox{ или дважды } = \drawAngle{E}$ \byref{prop:III.XX}. $\therefore \drawAngle{A} = \drawAngle{E}$. \end{center} @@ -6419,7 +6418,7 @@ \chapter*{Определения} и $\drawAngle{DAC} = \drawAngle{DBC}$. Добавим к~каждому \drawAngle{BCA,ACD},\\ -$\drawAngle{DAC,CAB} + \drawAngle{BCA,ACD} = \drawAngle{BCA,ACD} + \drawAngle{CDB} + \drawAngle{DBC} = \drawTwoRightAngles$ \bycref{prop:I.XXXII}. +$\drawAngle{DAC,CAB} + \drawAngle{BCA,ACD} = \drawAngle{BCA,ACD} + \drawAngle{CDB} + \drawAngle{DBC} = \drawTwoRightAngles$ \byref{prop:I.XXXII}. Так же можно показать, что\\ $\drawAngle{CDB,BDA} + \drawAngle{ABD,DBC} = \drawTwoRightAngles$. @@ -6481,9 +6480,9 @@ \chapter*{Определения} Проведем \drawUnitLine{BC} и~\drawUnitLine{BD}. Поскольку сегменты подобны,\\ -$\drawAngle{C} = \drawAngle{D}$ \bycref{def:III.XI}. +$\drawAngle{C} = \drawAngle{D}$ \byref{def:III.XI}. -Но $\drawAngle{C} > \drawAngle{D}$ \bycref{prop:I.XVI}, что невозможно. +Но $\drawAngle{C} > \drawAngle{D}$ \byref{prop:I.XVI}, что невозможно. Следовательно, никакая точка одного сегмента не расположена вовне другого сегмента, и~значит сегменты совпадают. \end{center} @@ -6552,7 +6551,7 @@ \chapter*{Определения} \drawUnitLine{CD} будет полностью совпадать с~\drawUnitLine{AB}. \end{center} -\noindent Равные сегменты на одной прямой и~по одну ее сторону также совпадают \bycref{prop:III.XXIII} и, следовательно, равны. +\noindent Равные сегменты на одной прямой и~по одну ее сторону также совпадают \byref{prop:III.XXIII} и, следовательно, равны. \qed @@ -6592,7 +6591,7 @@ \chapter*{Определения} Где они пересекаются, там и~находится центр круга. \end{center} -Поскольку \drawUnitLine{BC}, кончающаяся на окружности, рассекается перпендикуляром \drawUnitLine{EF}, который проходит через центр \bycref{prop:III.I}, так же и~\drawUnitLine{DF} проходит через центр, следовательно, их пересечение и~есть центр. +Поскольку \drawUnitLine{BC}, кончающаяся на окружности, рассекается перпендикуляром \drawUnitLine{EF}, который проходит через центр \byref{prop:III.I}, так же и~\drawUnitLine{DF} проходит через центр, следовательно, их пересечение и~есть центр. \qed @@ -6658,9 +6657,9 @@ \chapter*{Определения} $\drawUnitLine{BG} = \drawUnitLine{CG} = \drawUnitLine{EH} = \drawUnitLine{FH}$\\ и $\drawAngle{G} = \drawAngle{H}$. -$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$ \byref{prop:I.IV}. -Но $\drawAngle{A} = \drawAngle{D}$ \bycref{prop:III.XX}. +Но $\drawAngle{A} = \drawAngle{D}$ \byref{prop:III.XX}. $\therefore$ \drawFromCurrentPicture{ @@ -6678,8 +6677,8 @@ \chapter*{Определения} draw byLabelsOnCircle(E, F)(H); stopTempScale; } -подобны \bycref{def:III.XI}\\ -и равны \bycref{prop:III.XXIV}. +подобны \byref{def:III.XI}\\ +и равны \byref{prop:III.XXIV}. А если равные сегменты вычесть из равных кругов, оставшиеся сегменты будут равны,\\ то есть $ @@ -6692,7 +6691,7 @@ \chapter*{Определения} draw byNamedFilledCircleSegment(H); draw byNamedArcLabel(Hb); } -$ \bycref{ax:I.III}. +$ \byref{ax:I.III}. И $\therefore \drawArc{Gb} = \drawArc{Hb}$. \end{center} @@ -6765,9 +6764,9 @@ \chapter*{Определения} Cделаем $\drawAngle{BAK,KAC} = \drawAngle{D}$. -$\therefore \drawArc{GbI,GbII} = \drawArc{Hb}$ \bycref{prop:III.XXVI}. +$\therefore \drawArc{GbI,GbII} = \drawArc{Hb}$ \byref{prop:III.XXVI}. -Но $\drawArc{GbI} = \drawArc{Hb}$ \bycref{\hypref} +Но $\drawArc{GbI} = \drawArc{Hb}$ \byref{\hypref} $\therefore \drawArc{GbI} = \drawArc{GbI,GbII}$ часть равны целому, что невозможно. @@ -6826,13 +6825,13 @@ \chapter*{Определения} Поскольку $\circleK = \circleL$,\\ $\drawUnitLine{AK}, \drawUnitLine{BK} = \drawUnitLine{DL}, \drawUnitLine{EL}$,\\ -а также $\drawUnitLine{AB} = \drawUnitLine{DE}$ \bycref{\hypref}. +а также $\drawUnitLine{AB} = \drawUnitLine{DE}$ \byref{\hypref}. $\therefore \drawAngle{K} = \drawAngle{L}$. -$\therefore \drawArc{Kb} = \drawArc{Lb}$ \bycref{prop:III.XXVI}. +$\therefore \drawArc{Kb} = \drawArc{Lb}$ \byref{prop:III.XXVI}. -И $\therefore \drawArc[bottom][1/3]{K} = \drawArc[bottom][1/3]{L}$ \bycref{ax:I.III}. +И $\therefore \drawArc[bottom][1/3]{K} = \drawArc[bottom][1/3]{L}$ \byref{ax:I.III}. \end{center} \qed @@ -6885,12 +6884,12 @@ \chapter*{Определения} пусть \drawUnitLine{AK}, \drawUnitLine{BK} и~\drawUnitLine{DL}, \drawUnitLine{EL}\\ проведены из центров. -Поскольку $\drawArc{Kb} = \drawArc{Lb}$ \bycref{\hypref},\\ -$\drawAngle{K} = \drawAngle{L}$ \bycref{prop:III.XXVII}. +Поскольку $\drawArc{Kb} = \drawArc{Lb}$ \byref{\hypref},\\ +$\drawAngle{K} = \drawAngle{L}$ \byref{prop:III.XXVII}. Но $\drawUnitLine{AK} \mbox{ и~} \drawUnitLine{BK} = \drawUnitLine{DL} \mbox{ и~} \drawUnitLine{EL}$. -$\therefore \drawUnitLine{AB} = \drawUnitLine{DE}$ \bycref{prop:I.IV},\\ +$\therefore \drawUnitLine{AB} = \drawUnitLine{DE}$ \byref{prop:I.IV},\\ а это и~есть хорды, стягивающие равные дуги. \end{center} @@ -6932,18 +6931,18 @@ \chapter*{Определения} \begin{center} Проведем \drawUnitLine{AC,CB}. -Сделаем $\drawUnitLine{AC} = \drawUnitLine{CB}$ \bycref{prop:I.X}. +Сделаем $\drawUnitLine{AC} = \drawUnitLine{CB}$ \byref{prop:I.X}. -Проведем $\drawUnitLine{DC} \perp \drawUnitLine{AC,CB}$ \bycref{prop:I.XI}, она и~будет рассекать дугу. +Проведем $\drawUnitLine{DC} \perp \drawUnitLine{AC,CB}$ \byref{prop:I.XI}, она и~будет рассекать дугу. Проведем \drawUnitLine{DA} и~\drawUnitLine{DB}. Тогда у \drawLine[bottom]{AD,DC,CA} и \drawLine[bottom]{DB,BC,CD}\\ -$\drawUnitLine{AC} = \drawUnitLine{CB}$ \bycref{\constref},\\ +$\drawUnitLine{AC} = \drawUnitLine{CB}$ \byref{\constref},\\ \drawUnitLine{DC} общая\\ -и $\drawAngle{ACD} = \drawAngle{DCB}$ \bycref{\constref}. +и $\drawAngle{ACD} = \drawAngle{DCB}$ \byref{\constref}. -$\therefore \drawUnitLine{DA} = \drawUnitLine{DB}$ \bycref{prop:I.IV}\\ +$\therefore \drawUnitLine{DA} = \drawUnitLine{DB}$ \byref{prop:I.IV}\\ и $ \drawFromCurrentPicture{ startGlobalRotation(-arcAngle.El); @@ -6960,7 +6959,7 @@ \chapter*{Определения} stopAutoLabeling; stopGlobalRotation; } -$ \bycref{prop:III.XXVIII},\\ +$ \byref{prop:III.XXVIII},\\ и значит данная дуга рассечена. \end{center} @@ -6998,8 +6997,8 @@ \chapter*{Определения} Проведем \drawUnitLine{AE} и~\drawUnitLine{BE,EC}. -$\drawAngle{B}=\drawAngle{EAB}$ и~$\drawAngle{C} = \drawAngle{CAE}$ \bycref{prop:I.V},\\ -$\drawAngle{C} + \drawAngle{B} = \drawAngle{EAB,CAE} = \mbox{ половина } \drawTwoRightAngles= \drawRightAngle$ \bycref{prop:I.XXXII}. +$\drawAngle{B}=\drawAngle{EAB}$ и~$\drawAngle{C} = \drawAngle{CAE}$ \byref{prop:I.V},\\ +$\drawAngle{C} + \drawAngle{B} = \drawAngle{EAB,CAE} = \mbox{ половина } \drawTwoRightAngles= \drawRightAngle$ \byref{prop:I.XXXII}. \end{center} \defineNewPicture{ @@ -7062,7 +7061,7 @@ \chapter*{Определения} Возьмем на противоположной стороне окружности любую точку, к~которой проведем \drawUnitLine{BC} и~\drawUnitLine{AB}. -Поскольку $\drawAngle{B} + \drawAngle{D} = \drawTwoRightAngles$ \bycref{prop:III.XXII},\\ +Поскольку $\drawAngle{B} + \drawAngle{D} = \drawTwoRightAngles$ \byref{prop:III.XXII},\\ Но $\drawAngle{B} < \drawRightAngle$ (случай II.),\\ $\therefore$ \drawAngle{D} тупой. \end{center} @@ -7102,19 +7101,19 @@ \chapter*{Определения} \drawCurrentPictureInMargin \problem[3]{Е}{сли}{прямая \drawUnitLine{EF} касается круга, и~из точки касания проведена прямая \drawUnitLine{DB}, секущая круг, угол \drawAngle{FBD} между этой прямой и~касательной равен углу \drawAngle{A} в~накрестлежащем сегменте круга.} -Если хорда проходит через центр, то очевидно, что углы равны, поскольку и~тот и~тот прямые. \bycref{prop:III.XVI,prop:III.XXXI}. +Если хорда проходит через центр, то очевидно, что углы равны, поскольку и~тот и~тот прямые. \byref{prop:III.XVI,prop:III.XXXI}. -Если же нет, проведем $\drawUnitLine{AB} \perp \drawUnitLine{EF}$ из точки касания, которая будет проходить через центр \bycref{prop:III.XIX}. +Если же нет, проведем $\drawUnitLine{AB} \perp \drawUnitLine{EF}$ из точки касания, которая будет проходить через центр \byref{prop:III.XIX}. \begin{center} -$\therefore \drawAngle{ADB} = \drawAngle{ABE} $ \bycref{prop:III.XXXI}, -$\drawAngle{A} + \drawAngle{DBA} = \drawAngle{ABE} = \drawAngle{FBA}$ \bycref{prop:I.XXXII}. +$\therefore \drawAngle{ADB} = \drawAngle{ABE} $ \byref{prop:III.XXXI}, +$\drawAngle{A} + \drawAngle{DBA} = \drawAngle{ABE} = \drawAngle{FBA}$ \byref{prop:I.XXXII}. -$\therefore \drawAngle{A} = \drawAngle{FBD}$ \bycref{ax:I.III}. +$\therefore \drawAngle{A} = \drawAngle{FBD}$ \byref{ax:I.III}. -Теперь $\drawAngle{B} = \drawTwoRightAngles = \drawAngle{A} + \drawAngle{C}$ \bycref{prop:III.XXII}. +Теперь $\drawAngle{B} = \drawTwoRightAngles = \drawAngle{A} + \drawAngle{C}$ \byref{prop:III.XXII}. -$\therefore \drawAngle{DBE} = \drawAngle{C}$ \bycref{ax:I.III}, который и~будет углом в~накрестлежащем сегменте. +$\therefore \drawAngle{DBE} = \drawAngle{C}$ \byref{ax:I.III}, который и~будет углом в~накрестлежащем сегменте. \end{center} \qed @@ -7155,7 +7154,7 @@ \chapter*{Определения} \drawCurrentPictureInMargin \problem{Н}{а}{данной прямой \drawUnitLine{AB} описать сегмент круга, вмещающий угол, равный данному \drawAngle{givenRight}, \drawAngle{givenObtuse}, \drawAngle{givenAcute}.} -Если данный угол прямой, то рассечем отрезок и~опишем на нем полукруг, который, очевидно, будет содержать прямой угол \bycref{prop:III.XXXI}. +Если данный угол прямой, то рассечем отрезок и~опишем на нем полукруг, который, очевидно, будет содержать прямой угол \byref{prop:III.XXXI}. \begin{center} Если же данный угол острый или тупой, сделаем с~данной прямой на одном из ее концов $\drawAngle{DAB} = \drawAngle{givenAcute}$. @@ -7165,9 +7164,9 @@ \chapter*{Определения} Опишем \drawCircle[middle][1/5]{G} с~\drawUnitLine{AG} или \drawUnitLine{GB} в~качестве радиуса, ведь они равны. -\drawUnitLine{KD} касается \circleG\ \bycref{prop:III.XVI}. +\drawUnitLine{KD} касается \circleG\ \byref{prop:III.XVI}. -$\therefore$ \drawUnitLine{AB} делит круг на два сегмента, вмещающие углы, равные \drawAngle{EAK,BAE} и~\drawAngle{DAB}, соответственно равные \drawAngle{givenObtuse} и~\drawAngle{givenAcute} \bycref{prop:III.XXXII}. +$\therefore$ \drawUnitLine{AB} делит круг на два сегмента, вмещающие углы, равные \drawAngle{EAK,BAE} и~\drawAngle{DAB}, соответственно равные \drawAngle{givenObtuse} и~\drawAngle{givenAcute} \byref{prop:III.XXXII}. \end{center} \qed @@ -7199,7 +7198,7 @@ \chapter*{Определения} \problem{О}{т}{данного круга \drawCircle{O} отсечь сегмент, вмещающий данный прямолинейный угол \drawAngle{givenAngle}.} \begin{center} -Проведем \drawUnitLine{EF} \bycref{prop:III.XVII}, касательную к~кругу в~любой точке. +Проведем \drawUnitLine{EF} \byref{prop:III.XVII}, касательную к~кругу в~любой точке. В точке касания сделаем $\drawAngle{B} = \drawAngle{givenAngle}$, данному углу. @@ -7210,9 +7209,9 @@ \chapter*{Определения} Поскольку \drawUnitLine{EF} касательная,\\ и \drawUnitLine{AB} сечет ее,\\ -угол в~$\segmentO\ = \drawAngle{B}$ \bycref{prop:III.XXXII}. +угол в~$\segmentO\ = \drawAngle{B}$ \byref{prop:III.XXXII}. -Но $\drawAngle{B} = \drawAngle{givenAngle}$ \bycref{\constref} +Но $\drawAngle{B} = \drawAngle{givenAngle}$ \byref{\constref} \end{center} \qed @@ -7273,10 +7272,10 @@ \chapter*{Определения} \begin{center} Пусть \drawSizedLine{HC,EH,AE} проходит через центр, а~\drawSizedLine{BH,HD} нет, проведем \drawSizedLine{EB} и~\drawSizedLine{DE}. -Тогда $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EB}^2 - \drawSizedLine{EH}^2$ \bycref{prop:II.VI},\\ +Тогда $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EB}^2 - \drawSizedLine{EH}^2$ \byref{prop:II.VI},\\ или $\drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{HC,EH}^2 - \drawSizedLine{EH}^2$. -$\therefore \drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EH,AE} \times \drawSizedLine{AE}$ \bycref{prop:II.V}. +$\therefore \drawSizedLine{BH} \times \drawSizedLine{HD} = \drawSizedLine{EH,AE} \times \drawSizedLine{AE}$ \byref{prop:II.V}. \end{center} \defineNewPicture{ @@ -7348,10 +7347,10 @@ \chapter*{Определения} Проведем \drawSizedLine{BF} из центра к~точке касания. -$\drawSizedLine{DB}^2 = \drawSizedLine{CF,DC}^2 - \drawSizedLine{BF}^2$ \bycref{prop:I.XLVII},\\ +$\drawSizedLine{DB}^2 = \drawSizedLine{CF,DC}^2 - \drawSizedLine{BF}^2$ \byref{prop:I.XLVII},\\ или $\drawSizedLine{DB}^2 = \drawSizedLine{CF,DC}^2 - \drawSizedLine{CF}^2$. -$\therefore \drawSizedLine{DB}^2 = \drawSizedLine{FA,CF,DC} \times \drawSizedLine{DC}$ \bycref{prop:II.VI}. +$\therefore \drawSizedLine{DB}^2 = \drawSizedLine{FA,CF,DC} \times \drawSizedLine{DC}$ \byref{prop:II.VI}. \end{center} \defineNewPicture{ @@ -7385,11 +7384,11 @@ \chapter*{Определения} Если \drawSizedLine{CA,DC} не проходит через центр,\\ проведем \drawSizedLine{EA} и~\drawSizedLine{EC}. -Тогда $\drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DE}^2 - \drawSizedLine{EC}^2$ \bycref{prop:II.VI}. +Тогда $\drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DE}^2 - \drawSizedLine{EC}^2$ \byref{prop:II.VI}. То есть $\drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DE}^2 - \drawSizedLine{EB}^2$. -$\therefore \drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DB}^2$ \bycref{prop:III.XVIII}. +$\therefore \drawSizedLine{CA,DC} \times \drawSizedLine{CA} = \drawSizedLine{DB}^2$ \byref{prop:III.XVIII}. \end{center} \qed @@ -7433,8 +7432,8 @@ \chapter*{Определения} И проведем из центра \drawSizedLine{DF}, \drawSizedLine{BF} и~\drawSizedLine{EF}. -$\drawSizedLine{DE}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \bycref{prop:III.XXXVI},\\ -но $\drawSizedLine{DB}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \bycref{\hypref}. +$\drawSizedLine{DE}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \byref{prop:III.XXXVI},\\ +но $\drawSizedLine{DB}^2 = \drawSizedLine{CA,DC} \times \drawSizedLine{DC}$ \byref{\hypref}. $\therefore \drawSizedLine{DB}^2 = \drawSizedLine{DE}^2$,\\ и $\therefore \drawSizedLine{DB} = \drawSizedLine{DE}$. @@ -7442,11 +7441,11 @@ \chapter*{Определения} Тогда в~\drawLine{DB,DF,BF} и~\drawLine{EF,DF,DE}\\ $\drawSizedLine{BF} \mbox{ и~} \drawSizedLine{DB} = \drawSizedLine{EF} \mbox{ и~} \drawSizedLine{DE}$,\\ и \drawSizedLine{DF} общая обоим,\\ -$\therefore \drawAngle{B} = \drawAngle{E}$ \bycref{prop:I.VIII},\\ -но $\drawAngle{E} = \drawRightAngle$, прямому углу \bycref{prop:III.XVIII}. +$\therefore \drawAngle{B} = \drawAngle{E}$ \byref{prop:I.VIII},\\ +но $\drawAngle{E} = \drawRightAngle$, прямому углу \byref{prop:III.XVIII}. $\therefore \drawAngle{B} = \drawRightAngle$, прямому углу,\\ -и $\therefore$ \drawSizedLine{DB} касается круга \bycref{prop:III.XVI}. +и $\therefore$ \drawSizedLine{DB} касается круга \byref{prop:III.XVI}. \end{center} \qed @@ -7616,12 +7615,12 @@ \chapter*{Определения} и если $\drawUnitLine{EC,BE} = \drawUnitLine{GH}$, задача решена. Если же \drawUnitLine{EC,BE} не равна \drawUnitLine{GH},\\ -$\drawUnitLine{EC,BE} > \drawUnitLine{GH}$ \bycref{\hypref}. +$\drawUnitLine{EC,BE} > \drawUnitLine{GH}$ \byref{\hypref}. -Сделаем $\drawUnitLine{EC} = \drawUnitLine{GH}$ \bycref{prop:I.III},\\ +Сделаем $\drawUnitLine{EC} = \drawUnitLine{GH}$ \byref{prop:I.III},\\ взяв \drawUnitLine{EC} за радиус, построим \drawCircle{C}, секущий \circleO, \\ и проведем \drawUnitLine{AC}, которая и~будет искомой прямой,\\ -поскольку $\drawUnitLine{AC} = \drawUnitLine{EC} = \drawUnitLine{GH}$ \bycref{prop:I.XV,\constref}. +поскольку $\drawUnitLine{AC} = \drawUnitLine{EC} = \drawUnitLine{GH}$ \byref{prop:I.XV,\constref}. \end{center} \qed @@ -7668,21 +7667,21 @@ \chapter*{Определения} \problem{В}{ данный}{круг \drawCircle{O} вписать треугольник, равноугольный данному.} \begin{center} -Проведем касательную \drawUnitLine{GH} к~любой точке круга \bycref{prop:III.XVII}. +Проведем касательную \drawUnitLine{GH} к~любой точке круга \byref{prop:III.XVII}. -В точке касания сделаем $\drawAngle{CAH} = \drawAngle{E}$ \bycref{prop:I.XXIII}. +В точке касания сделаем $\drawAngle{CAH} = \drawAngle{E}$ \byref{prop:I.XXIII}. Так же сделаем $\drawAngle{GAB} = \drawAngle{F}$\\ и проведем \drawUnitLine{BC}. -Поскольку $\drawAngle{CAH} = \drawAngle{E}$ \bycref{\constref}\\ -и $\drawAngle{CAH} = \drawAngle{B}$ \bycref{prop:III.XXXII}. +Поскольку $\drawAngle{CAH} = \drawAngle{E}$ \byref{\constref}\\ +и $\drawAngle{CAH} = \drawAngle{B}$ \byref{prop:III.XXXII}. $\therefore \drawAngle{B} = \drawAngle{E}$. Также и~$\drawAngle{C} = \drawAngle{F}$ по той же причине. -$\therefore \drawAngle{BAC} = \drawAngle{D}$ \bycref{prop:I.XXXII}. +$\therefore \drawAngle{BAC} = \drawAngle{D}$ \byref{prop:I.XXXII}. \end{center} \noindent И, следовательно, вписанный в~круг треугольник равноуголен данному. @@ -7743,10 +7742,10 @@ \chapter*{Определения} Из центра данного круга проведем радиус \drawUnitLine{KC}. -Сделаем $\drawAngle{CKA} = \drawAngle{DFH}$ \bycref{prop:I.XXIII}\\ +Сделаем $\drawAngle{CKA} = \drawAngle{DFH}$ \byref{prop:I.XXIII}\\ и $\drawAngle{BKC} = \drawAngle{GED}$. -Через концы радиусов проведем \drawUnitLine{NL}, \drawUnitLine{LM} и~\drawUnitLine{MN}, касательные к~кругу \bycref{prop:III.XVII}. +Через концы радиусов проведем \drawUnitLine{NL}, \drawUnitLine{LM} и~\drawUnitLine{MN}, касательные к~кругу \byref{prop:III.XVII}. Четыре угла \drawFromCurrentPicture[bottom]{ @@ -7756,19 +7755,19 @@ \chapter*{Определения} draw byLabelsOnPolygon(L, A, K, C)(ALL_LABELS, 0); stopTempAngleScale; }, -взятые вместе, равны четырем прямым углам \bycref{prop:I.XXXII}. +взятые вместе, равны четырем прямым углам \byref{prop:I.XXXII}. -Но \drawAngle{C} и~\drawAngle{A} прямые \bycref{\constref}. +Но \drawAngle{C} и~\drawAngle{A} прямые \byref{\constref}. $\therefore \drawAngle{L} + \drawAngle{CKA} = \drawTwoRightAngles$, двум прямым углам. -Но $\drawAngle{DFH,DFE} = \drawTwoRightAngles$ \bycref{prop:I.XIII}\\ -и $\drawAngle{CKA} = \drawAngle{DFH}$ \bycref{\constref} и~$\therefore \drawAngle{L} = \drawAngle{DFE}$. +Но $\drawAngle{DFH,DFE} = \drawTwoRightAngles$ \byref{prop:I.XIII}\\ +и $\drawAngle{CKA} = \drawAngle{DFH}$ \byref{\constref} и~$\therefore \drawAngle{L} = \drawAngle{DFE}$. Таким же образом можно показать, что\\ $\drawAngle{N} = \drawAngle{FED}$. -$\therefore \drawAngle{M} = \drawAngle{D}$ \bycref{prop:I.XXXII}\\ +$\therefore \drawAngle{M} = \drawAngle{D}$ \byref{prop:I.XXXII}\\ и значит, описанный около круга треугольник равноуголен данному. \end{center} @@ -7819,7 +7818,7 @@ \chapter*{Определения} stopTempScale; } вписать круг.} -Рассечем пополам \drawAngle{EBD,DBF} и~\drawAngle{GCD,DCF} \bycref{prop:I.IX} с~помощью \drawUnitLine{BD} и~\drawUnitLine{CD}. +Рассечем пополам \drawAngle{EBD,DBF} и~\drawAngle{GCD,DCF} \byref{prop:I.IX} с~помощью \drawUnitLine{BD} и~\drawUnitLine{CD}. Из точки, где они встречаются, проведем \drawUnitLine{FD}, \drawUnitLine{ED} и~\drawUnitLine{GD} соотвтетственно перпендикулярные \drawUnitLine{BC}, \drawUnitLine{AB} и~\drawUnitLine{CA}. @@ -7844,14 +7843,14 @@ \chapter*{Определения} }\\ $\drawAngle{DBF} = \drawAngle{EBD}$, $\drawAngle{F} = \drawAngle{E}$ и~\drawUnitLine{BD} общая. -$\therefore \drawUnitLine{ED} = \drawUnitLine{FD}$ \bycref{prop:I.IV,prop:I.XXVI}. +$\therefore \drawUnitLine{ED} = \drawUnitLine{FD}$ \byref{prop:I.IV,prop:I.XXVI}. Так же можно показать, что $\drawUnitLine{GD} = \drawUnitLine{FD}$. $\therefore \drawUnitLine{FD} = \drawUnitLine{ED} = \drawUnitLine{GD}$. \end{center} -\noindent Взяв любую из этих линий за радиус, опишем \drawCircle[middle][1/2]{D}, и~его окружность будет проходить через концы других двух, а~стороны данного треугольника, перпендикулярные к~трем радиусам и~проходящие через их концы, касаются круга \bycref{prop:III.XVI}, который, таким образом, вписан в~данный треугольник. +\noindent Взяв любую из этих линий за радиус, опишем \drawCircle[middle][1/2]{D}, и~его окружность будет проходить через концы других двух, а~стороны данного треугольника, перпендикулярные к~трем радиусам и~проходящие через их концы, касаются круга \byref{prop:III.XVI}, который, таким образом, вписан в~данный треугольник. \qed @@ -7902,10 +7901,10 @@ \chapter*{Определения} \begin{center} Сделаем $\drawUnitLine{AD} = \drawUnitLine{DB}$\\ -и~$\drawUnitLine{CE} = \drawUnitLine{EA}$ \bycref{prop:I.X}. +и~$\drawUnitLine{CE} = \drawUnitLine{EA}$ \byref{prop:I.X}. Из точек рассечения проведем\\ -\drawUnitLine{DF} и~\drawUnitLine{EF} $\perp$ \drawUnitLine{AD} и~\drawUnitLine{CE} \bycref{prop:I.XI},\\ +\drawUnitLine{DF} и~\drawUnitLine{EF} $\perp$ \drawUnitLine{AD} и~\drawUnitLine{CE} \byref{prop:I.XI},\\ из точки их пересечения проведем\\ \drawUnitLine{BF}, \drawUnitLine{AF} и~\drawUnitLine{CF}\\ и~опишем круг с~любой из них в~качестве радиуса, это и~будет искомый круг. @@ -7924,11 +7923,11 @@ \chapter*{Определения} draw byNamedLineSeq(0)(AF,DF,AD); stopAutoLabeling; }\\ -$\drawUnitLine{DB} = \drawUnitLine{AD}$ \bycref{\constref},\\ +$\drawUnitLine{DB} = \drawUnitLine{AD}$ \byref{\constref},\\ \drawUnitLine{DF} общая,\\ -$\drawAngle{FDB} = \drawAngle{ADF}$ \bycref{\constref}. +$\drawAngle{FDB} = \drawAngle{ADF}$ \byref{\constref}. -$\therefore \drawUnitLine{AF} = \drawUnitLine{BF}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{AF} = \drawUnitLine{BF}$ \byref{prop:I.IV}. Так же можно показать и~что $\drawUnitLine{CF} = \drawUnitLine{AF}$. @@ -7975,17 +7974,17 @@ \chapter*{Определения} проведем \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{DA} и~\drawUnitLine{CD},\\ тогда \drawLine[middle][squareABCD]{DA,CD,BC,AB} квадрат. -Поскольку \drawAngle{A} и~\drawAngle{D} каждый в~полукруге, оба являются прямыми \bycref{prop:III.XXXI}. +Поскольку \drawAngle{A} и~\drawAngle{D} каждый в~полукруге, оба являются прямыми \byref{prop:III.XXXI}. -$\therefore \drawUnitLine{CD} \parallel \drawUnitLine{AB}$ \bycref{prop:I.XXVIII}. +$\therefore \drawUnitLine{CD} \parallel \drawUnitLine{AB}$ \byref{prop:I.XXVIII}. так же и~$\drawUnitLine{DA} \parallel \drawUnitLine{BC}$. -И, поскольку $\drawAngle{AEB} = \drawAngle{DEA}$ \bycref{\constref},\\ -и $\drawUnitLine{BE} = \drawUnitLine{AE} = \drawUnitLine{DE}$ \bycref{def:I.XV}, -$\therefore \drawUnitLine{AB} = \drawUnitLine{DA}$ \bycref{prop:I.IV}. +И, поскольку $\drawAngle{AEB} = \drawAngle{DEA}$ \byref{\constref},\\ +и $\drawUnitLine{BE} = \drawUnitLine{AE} = \drawUnitLine{DE}$ \byref{def:I.XV}, +$\therefore \drawUnitLine{AB} = \drawUnitLine{DA}$ \byref{prop:I.IV}. -Поскольку смежные углы и~стороны параллелограмма \squareABCD\ равны, все его стороны и~углы равны \bycref{prop:I.XXXIV}. +Поскольку смежные углы и~стороны параллелограмма \squareABCD\ равны, все его стороны и~углы равны \byref{prop:I.XXXIV}. И~$\therefore$ \squareABCD, вписанный в~данный круг, является квадратом. \end{center} @@ -8031,15 +8030,15 @@ \chapter*{Определения} Проведем два перпендикулярных друг другу диаметра данного круга \drawUnitLine{BD}, \drawUnitLine{AC} и~через их концы проведем \drawUnitLine{HK}, \drawUnitLine{GH}, \drawUnitLine{FG} и~\drawUnitLine{KF}, касательные к~кругу,\\ тогда \drawLine[middle][squareFGHK]{KF,HK,GH,FG} будет квадратом. -$\drawAngle{C} = \drawRightAngle$ прямому углу \bycref{prop:III.XVIII},\\ -так же и~$\drawAngle{E} = \drawRightAngle$ \bycref{\constref},\\ +$\drawAngle{C} = \drawRightAngle$ прямому углу \byref{prop:III.XVIII},\\ +так же и~$\drawAngle{E} = \drawRightAngle$ \byref{\constref},\\ $\therefore \drawUnitLine{HK} \parallel \drawUnitLine{BD}$. Таким же образом можно показать, что $\drawUnitLine{FG} \parallel \drawUnitLine{BD}$,\\ а~также что $\drawUnitLine{GH} \mbox{ и~} \drawUnitLine{KF} \parallel \drawUnitLine{AC}$. $\therefore$ \squareFGHK\ параллелограмм,\\ -и поскольку $\drawAngle{G} = \drawAngle{F} = \drawAngle{K} = \drawAngle{H} = \drawAngle{C}$, все они прямые \bycref{prop:I.XXXIV}. +и поскольку $\drawAngle{G} = \drawAngle{F} = \drawAngle{K} = \drawAngle{H} = \drawAngle{C}$, все они прямые \byref{prop:I.XXXIV}. Также очевидно, что \drawUnitLine{HK}, \drawUnitLine{GH}, \drawUnitLine{FG} и~\drawUnitLine{KF} равны между собой. @@ -8087,21 +8086,21 @@ \chapter*{Определения} и $\drawUnitLine{KC} = \drawUnitLine{DK}$. Проведем $\drawUnitLine{FG,GK} \parallel \drawUnitLine{AE,ED}$\\ -и $\drawUnitLine{EG,GH} \parallel \drawUnitLine{DK,KC}$ \bycref{prop:I.XXXI}. +и $\drawUnitLine{EG,GH} \parallel \drawUnitLine{DK,KC}$ \byref{prop:I.XXXI}. $\therefore$ \drawPolygon{EDKG} параллелограмм. -И поскольку $\drawUnitLine{AE,ED} = \drawUnitLine{DK,KC}$ \bycref{\hypref},\\ +И поскольку $\drawUnitLine{AE,ED} = \drawUnitLine{DK,KC}$ \byref{\hypref},\\ $\drawUnitLine{ED} = \drawUnitLine{DK}$. -$\therefore$ \polygonEDKG\ равносторонний \bycref{prop:I.XXXIV}. +$\therefore$ \polygonEDKG\ равносторонний \byref{prop:I.XXXIV}. Таким же образом можно показать, что $ \drawPolygon{GKCH} = \drawPolygon{FGHB}$ равносторонние параллелограммы. $\therefore \drawUnitLine{AE} = \drawUnitLine{GK} = \drawUnitLine{GH} = \drawUnitLine{FG}$. \end{center} -И значит если построить круг с~центром в~точке схождения этих линий с~любой их них в~качестве радиуса, он будет вписан в~квадрат \bycref{prop:III.XVI}. +И значит если построить круг с~центром в~точке схождения этих линий с~любой их них в~качестве радиуса, он будет вписан в~квадрат \byref{prop:III.XVI}. \qed @@ -8138,7 +8137,7 @@ \chapter*{Определения} Проведем пересекающие друг друга диагонали \drawUnitLine{AE,EC} и~\drawUnitLine{BE,ED}. Тогда, $\because$ стороны у~\drawPolygon{ACD} и~\drawPolygon{ABC} равны,\\ а~основание \drawUnitLine{AE,EC} общее,\\ -$\drawAngle{DAE} = \drawAngle{EAB}$ \bycref{prop:I.VIII},\\ +$\drawAngle{DAE} = \drawAngle{EAB}$ \byref{prop:I.VIII},\\ то есть \drawAngle{DAE,EAB} рассечен пополам. Так же можно показать и~что \drawAngle{ABE,EBC} рассечен пополам. @@ -8147,7 +8146,7 @@ \chapter*{Определения} Значит, и~их половины $\drawAngle{EAB} = \drawAngle{ABE}$. -$\therefore \drawUnitLine{BE} = \drawUnitLine{AE}$ \bycref{prop:I.VI}. +$\therefore \drawUnitLine{BE} = \drawUnitLine{AE}$ \byref{prop:I.VI}. Так же можно показать,\\ что $\drawUnitLine{AE} = \drawUnitLine{BE} = \drawUnitLine{EC} = \drawUnitLine{ED}$. @@ -8196,33 +8195,33 @@ \chapter*{Определения} \begin{center} Возьмем любую прямую \drawProportionalLine{AC,CB} \\ -и разделим ее так, что $\drawProportionalLine{AC,CB} \times \drawProportionalLine{CB} = \drawProportionalLine{AC}^2$ \bycref{prop:II.XI}. +и разделим ее так, что $\drawProportionalLine{AC,CB} \times \drawProportionalLine{CB} = \drawProportionalLine{AC}^2$ \byref{prop:II.XI}. -С \drawProportionalLine{AC,CB} в~качестве радиуса опишем \drawCircle[middle][1/5]{A} и~впишем в~него от конца радиуса $\drawProportionalLine{BD} = \drawProportionalLine{AC}$ \bycref{prop:IV.I}. +С \drawProportionalLine{AC,CB} в~качестве радиуса опишем \drawCircle[middle][1/5]{A} и~впишем в~него от конца радиуса $\drawProportionalLine{BD} = \drawProportionalLine{AC}$ \byref{prop:IV.I}. Проведем \drawProportionalLine{AD}. Тогда \drawLine{AD,BD,CB,AC} и~есть искомый треугольник. -Действительно, проведем \drawProportionalLine{CD} и~опишем \drawCircle[middle][1/3]{F} около \drawLine{AD,CD,AC} \bycref{prop:IV.V}. +Действительно, проведем \drawProportionalLine{CD} и~опишем \drawCircle[middle][1/3]{F} около \drawLine{AD,CD,AC} \byref{prop:IV.V}. Поскольку $\drawProportionalLine{AC,CB} \times \drawProportionalLine{CB} = \drawProportionalLine{AC}^2 = \drawProportionalLine{BD}^2$,\\ -$\therefore$ \drawProportionalLine{BD} касается \circleF\ \bycref{prop:III.XXXVII},\\ -$\therefore \drawAngle{BDC} = \drawAngle{A}$ \bycref{prop:III.XXXII}. +$\therefore$ \drawProportionalLine{BD} касается \circleF\ \byref{prop:III.XXXVII},\\ +$\therefore \drawAngle{BDC} = \drawAngle{A}$ \byref{prop:III.XXXII}. Добавим \drawAngle{CDA} к~каждому,\\ $\therefore \drawAngle{BDC} + \drawAngle{CDA} = \drawAngle{A} + \drawAngle{CDA}$. -Но $\drawAngle{BDC} + \drawAngle{CDA} \mbox{ или } \drawAngle{BDC,CDA} = \drawAngle{B}$ \bycref{prop:I.V}. +Но $\drawAngle{BDC} + \drawAngle{CDA} \mbox{ или } \drawAngle{BDC,CDA} = \drawAngle{B}$ \byref{prop:I.V}. -Поскольку $\drawProportionalLine{AD} = \drawProportionalLine{AC,CB}$ \bycref{prop:I.V},\\ -$\therefore \drawAngle{B} = \drawAngle{A} + \drawAngle{CDA} = \drawAngle{C}$ \bycref{prop:I.XXXII}. +Поскольку $\drawProportionalLine{AD} = \drawProportionalLine{AC,CB}$ \byref{prop:I.V},\\ +$\therefore \drawAngle{B} = \drawAngle{A} + \drawAngle{CDA} = \drawAngle{C}$ \byref{prop:I.XXXII}. -$\therefore \drawProportionalLine{CD} = \drawProportionalLine{BD}$ \bycref{prop:I.VI}. +$\therefore \drawProportionalLine{CD} = \drawProportionalLine{BD}$ \byref{prop:I.VI}. -$\therefore \drawProportionalLine{BD} = \drawProportionalLine{AC} = \drawProportionalLine{CD}$ \bycref{\constref}. +$\therefore \drawProportionalLine{BD} = \drawProportionalLine{AC} = \drawProportionalLine{CD}$ \byref{\constref}. -$\therefore \drawAngle{A} = \drawAngle{CDA}$ \bycref{prop:I.V}. +$\therefore \drawAngle{A} = \drawAngle{CDA}$ \byref{prop:I.V}. $\therefore \drawAngle{B} = \drawAngle{BDC,CDA} = \drawAngle{C} = \drawAngle{A} + \drawAngle{CDA} = \mbox{ дважды } \drawAngle{A}$. @@ -8270,22 +8269,22 @@ \chapter*{Определения} \problem[2]{В}{ данный}{круг \drawCircle[middle][1/6]{O} вписать равносторонний и~равноугольный пятиугольник.} \begin{center} -Построим равнобедренный треугольник, в~котором каждый из углов при основании вдвое больше угла в~вершине \bycref{prop:IV.X}. +Построим равнобедренный треугольник, в~котором каждый из углов при основании вдвое больше угла в~вершине \byref{prop:IV.X}. Впишем равноугольный ему\\ -треугольник \drawPolygon{ACD} в~данный круг \drawCircle[middle][1/6]{O} \bycref{prop:IV.II}. +треугольник \drawPolygon{ACD} в~данный круг \drawCircle[middle][1/6]{O} \byref{prop:IV.II}. -Рассечем \drawAngle{ECA,DCE} и~\drawAngle{ADB,BDC} пополам \bycref{prop:I.IX}. +Рассечем \drawAngle{ECA,DCE} и~\drawAngle{ADB,BDC} пополам \byref{prop:I.IX}. Проведем \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{EA} и~\drawUnitLine{DE}. Поскольку углы \drawAngle{ECA}, \drawAngle{DCE}, \drawAngle{CAD}, \drawAngle{BDC} и~\drawAngle{ADB}\\ -равны между собой, дуги, на которых они стоят, тоже равны \bycref{prop:III.XXVI}. +равны между собой, дуги, на которых они стоят, тоже равны \byref{prop:III.XXVI}. -И~$\therefore$ \drawUnitLine{CD}, \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{EA} и~\drawUnitLine{DE}, стягивающие эти дуги, также равны между собой \bycref{prop:III.XXIX}. +И~$\therefore$ \drawUnitLine{CD}, \drawUnitLine{BC}, \drawUnitLine{AB}, \drawUnitLine{EA} и~\drawUnitLine{DE}, стягивающие эти дуги, также равны между собой \byref{prop:III.XXIX}. И~$\therefore$ пятиугольник является равносторонним,\\ -а~также равноугольным, поскольку все его углы стоят на равных дугах \bycref{prop:III.XXVII}. +а~также равноугольным, поскольку все его углы стоят на равных дугах \byref{prop:III.XXVII}. \end{center} \qed @@ -8338,7 +8337,7 @@ \chapter*{Определения} \problem{О}{коло}{данного круга \drawCircle[middle][1/4]{F} описать равносторонний и~равноугольный пятиугольник.} \begin{center} -Проведем пять касательных через вершины любого правильного пятиугольника, вписанного в~круг \circleF\ \bycref{prop:III.XVII}. +Проведем пять касательных через вершины любого правильного пятиугольника, вписанного в~круг \circleF\ \byref{prop:III.XVII}. Эти пять касательных и~будут образовывать искомый пятиугольник. @@ -8351,17 +8350,17 @@ \chapter*{Определения} \nointerlineskip\hbox{\drawProportionalLine{FD}} }\right.$.\\ В \drawLine{FB,FK,BK} и~\drawLine{FK,FC,KC}\\ -$\drawProportionalLine{BK} = \drawProportionalLine{KC}$ \bycref{prop:I.XLVII},\\ +$\drawProportionalLine{BK} = \drawProportionalLine{KC}$ \byref{prop:I.XLVII},\\ $\drawProportionalLine{FC} = \drawProportionalLine{FB}$ и~\drawProportionalLine{FK} общая. -$\therefore \drawAngle{FKB} = \drawAngle{CKF}$ и~$\therefore \drawAngle{BFK} = \drawAngle{KFC}$ \bycref{prop:I.VIII}. +$\therefore \drawAngle{FKB} = \drawAngle{CKF}$ и~$\therefore \drawAngle{BFK} = \drawAngle{KFC}$ \byref{prop:I.VIII}. $\therefore \drawAngle{FKB,CKF} = \mbox{ дважды } \drawAngle{CKF}$, и~$\drawAngle{BFK,KFC} = \mbox{ дважды } \drawAngle{KFC}$. Таким же образом можно показать, что\\ $\drawAngle{FLC,DLF} = \mbox{ дважды } \drawAngle{FLC}$, и~$\drawAngle{CFL,LFD} = \mbox{ дважды } \drawAngle{CFL}$. -Но $\drawAngle{BFK,KFC} = \drawAngle{CFL,LFD}$ \bycref{prop:III.XXVII},\\ +Но $\drawAngle{BFK,KFC} = \drawAngle{CFL,LFD}$ \byref{prop:III.XXVII},\\ $\therefore$ и~их половины $\drawAngle{KFC} = \drawAngle{CFL}$, а~также $\drawAngle{FCK} = \drawAngle{LCF}$\\ и \drawProportionalLine{FC} общая. @@ -8445,7 +8444,7 @@ \chapter*{Определения} будет данным равноугольным и~равносторонним пятиугольником, в~который требуется вписать круг. \begin{center} -Сделаем $\drawAngle{HCF} = \drawAngle{FCK}$ и~$\drawAngle{FDE} = \drawAngle{KDF}$ \bycref{prop:I.IX}. +Сделаем $\drawAngle{HCF} = \drawAngle{FCK}$ и~$\drawAngle{FDE} = \drawAngle{KDF}$ \byref{prop:I.IX}. Проведем \drawUnitLine{FD}, \drawUnitLine{FC}, \drawUnitLine{FB}, \drawUnitLine{FE}, и~т. д. @@ -8468,7 +8467,7 @@ \chapter*{Определения} stopTempScale; }\\ $\drawUnitLine{KD,CK} = \drawUnitLine{HC,BH}$, $\drawAngle{HCF} = \drawAngle{FCK}$ и~\drawUnitLine{FC} общая обоим,\\ -$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ и~$\drawAngle{FBH} = \drawAngle{KDF}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ и~$\drawAngle{FBH} = \drawAngle{KDF}$ \byref{prop:I.IV}. И поскольку $\drawAngle{B} = \drawAngle{D} = \mbox{ дважды } \drawAngle{KDF}$,\\ $\therefore \mbox{ дважды } \drawAngle{FBH}$, а~значит \drawAngle{B} рассечен \drawUnitLine{FB} пополам. @@ -8495,15 +8494,15 @@ \chapter*{Определения} stopAutoLabeling; stopTempScale; }\\ -получим $\drawAngle{HCF} = \drawAngle{FCK}$ \bycref{\constref}, \drawUnitLine{FC} общая\\ +получим $\drawAngle{HCF} = \drawAngle{FCK}$ \byref{\constref}, \drawUnitLine{FC} общая\\ и $\drawAngle{H} = \drawAngle{K} = \drawRightAngle \mbox{ прямому углу}$. -$\therefore \drawUnitLine{FK} = \drawUnitLine{FH}$ \bycref{prop:I.XXVI}. +$\therefore \drawUnitLine{FK} = \drawUnitLine{FH}$ \byref{prop:I.XXVI}. \end{center} Так же можно показать, что пять перпендикуляров к~сторонам пятиугольника равны между собой. -Опишем \drawCircle[middle][1/5]{F} с~одним из перпендикуляров в~качестве радиуса, это и~будет искомый вписанный круг. Поскольку если он не касается сторон пятиугольника, но сечет их, то прямая, проведенная через конец диаметра под прямым углом, будет проходить внутри круга, что, как было показано \bycref{prop:III.XVI}, невозможно. +Опишем \drawCircle[middle][1/5]{F} с~одним из перпендикуляров в~качестве радиуса, это и~будет искомый вписанный круг. Поскольку если он не касается сторон пятиугольника, но сечет их, то прямая, проведенная через конец диаметра под прямым углом, будет проходить внутри круга, что, как было показано \byref{prop:III.XVI}, невозможно. \qed @@ -8548,7 +8547,7 @@ \chapter*{Определения} Рассечем пополам \drawAngle{BCF,FCD} и~\drawAngle{CDF,FDE} с~помощью \drawUnitLine{FC} и~\drawUnitLine{FD} и~из точки их пересечения проведем \drawUnitLine{FE}, \drawUnitLine{FA} и~\drawUnitLine{FB}. $\drawAngle{BCF,FCD} = \drawAngle{CDF,FDE}$, $\drawAngle{FCD} = \drawAngle{CDF}$,\\ -$\therefore \drawUnitLine{FD} = \drawUnitLine{FC}$ \bycref{prop:I.VI}. +$\therefore \drawUnitLine{FD} = \drawUnitLine{FC}$ \byref{prop:I.VI}. И поскольку в \drawFromCurrentPicture{ @@ -8566,7 +8565,7 @@ \chapter*{Определения} }\\ $\drawUnitLine{BC} = \drawUnitLine{CD}$ и~\drawUnitLine{FC} общая,\\ а~также $\drawAngle{BCF} = \drawAngle{FCD}$,\\ -$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ \bycref{prop:I.IV}. +$\therefore \drawUnitLine{FB} = \drawUnitLine{FD}$ \byref{prop:I.IV}. Так же можно показать, что\\ $\drawUnitLine{FA} = \drawUnitLine{FE} = \drawUnitLine{FB}$. @@ -8642,11 +8641,11 @@ \chapter*{Определения} stopAutoLabeling; } равносторонние,\\ -так как $\drawAngle{DGC} = \drawAngle{EGD} = \dfrac{1}{3} \drawTwoRightAngles$ \bycref{prop:I.XXXII}, но $\drawAngle{DGC,EGD,FGE} = \drawTwoRightAngles$ \bycref{prop:I.XIII}. +так как $\drawAngle{DGC} = \drawAngle{EGD} = \dfrac{1}{3} \drawTwoRightAngles$ \byref{prop:I.XXXII}, но $\drawAngle{DGC,EGD,FGE} = \drawTwoRightAngles$ \byref{prop:I.XIII}. -$\therefore \drawAngle{DGC} = \drawAngle{EGD} = \drawAngle{FGE} = \dfrac{1}{3} \drawTwoRightAngles$ \bycref{prop:I.XXXII}, и~вертикальные углы равны между собой \bycref{prop:I.XV}\\ -и~стоят на равных дугах \bycref{prop:III.XXVI},\\ -которые стагивают равные хорды \bycref{prop:III.XXIX}. +$\therefore \drawAngle{DGC} = \drawAngle{EGD} = \drawAngle{FGE} = \dfrac{1}{3} \drawTwoRightAngles$ \byref{prop:I.XXXII}, и~вертикальные углы равны между собой \byref{prop:I.XV}\\ +и~стоят на равных дугах \byref{prop:III.XXVI},\\ +которые стагивают равные хорды \byref{prop:III.XXIX}. И поскольку каждый из углов шестиугольника вдвое больше угла равностороннего треугольника, этот шестиугольник также равносторонний. \end{center} @@ -8724,7 +8723,7 @@ \chapter*{Определения} $\therefore$ дуга, стягиваемая $\drawUnitLine{FC}$, $= \dfrac{1}{15}$ всей окружности. \end{center} -А значит, если прямые, равные \drawUnitLine{FC}, вписать в~круг \bycref{prop:IV.I}, получим равносторонний и~равноугольный пятнадцатиугольник, вписанный в~круг. +А значит, если прямые, равные \drawUnitLine{FC}, вписать в~круг \byref{prop:IV.I}, получим равносторонний и~равноугольный пятнадцатиугольник, вписанный в~круг. \qed @@ -9090,17 +9089,17 @@ \chapter*{Аксиомы} Пусть $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$,\\ тогда $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$,\\ всякое равнократное $3\drawMagnitude{I}$ и~$3\drawMagnitude{III}$ будет равнократным \drawMagnitude{I} и~\drawMagnitude{III},\\ -и всякое равнократное $2\drawMagnitude{II}$ и~$2\drawMagnitude{IV}$ будет равнократным \drawMagnitude{II} и~\drawMagnitude{IV} \bycref{prop:V.III}. +и всякое равнократное $2\drawMagnitude{II}$ и~$2\drawMagnitude{IV}$ будет равнократным \drawMagnitude{II} и~\drawMagnitude{IV} \byref{prop:V.III}. То есть $M$ раз $3\drawMagnitude{I}$ и~$M$ раз $3\drawMagnitude{III}$\\ равнократны \drawMagnitude{I} и~\drawMagnitude{III},\\ и $m$ раз $2\drawMagnitude{II}$ и~$m$ раз $2\drawMagnitude{IV}$\\ равнократны $2\drawMagnitude{II}$ и~$2\drawMagnitude{IV}$. -Но $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$ \bycref{\hypref},\\ +Но $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$ \byref{\hypref},\\ $\therefore$ если $M 3\drawMagnitude{I} <, =, \mbox{ или } > m 2 \drawMagnitude{II}$,\\ -то $M 3\drawMagnitude{III} <, =, \mbox{ или } > m 2 \drawMagnitude{IV}$ \bycref{def:V.V},\\ -и значит $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$ \bycref{def:V.V}. +то $M 3\drawMagnitude{III} <, =, \mbox{ или } > m 2 \drawMagnitude{IV}$ \byref{def:V.V},\\ +и значит $3\drawMagnitude{I} : 2\drawMagnitude{II} :: 3\drawMagnitude{III} : 2\drawMagnitude{IV}$ \byref{def:V.V}. \end{center} Те же соображения работают для любых равнократных первой и~третьей величин и~любых равнократных второй и~четвертой. @@ -9181,7 +9180,7 @@ \chapter*{Аксиомы} \begin{center} Пусть $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$. -Следовательно \bycref{def:V.V}, если $\drawMagnitude{dI} > \drawMagnitude{dII}$, то $\drawMagnitude{dIII} > \drawMagnitude{dIV}$. +Следовательно \byref{def:V.V}, если $\drawMagnitude{dI} > \drawMagnitude{dII}$, то $\drawMagnitude{dIII} > \drawMagnitude{dIV}$. Но если $\drawMagnitude{I} > \drawMagnitude{II}$,\\ то $\drawMagnitude{dI} > \drawMagnitude{dII}$ и~$\drawMagnitude{dIII} > \drawMagnitude{dIV}$,\\ @@ -9218,7 +9217,7 @@ \chapter*{Аксиомы} Пусть $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$,\\ тогда, перевернув, получим $\magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$. -Если $M \magnitudeI < m \magnitudeII$, то $M \magnitudeIII < m \magnitudeIV$ \bycref{def:V.V}. +Если $M \magnitudeI < m \magnitudeII$, то $M \magnitudeIII < m \magnitudeIV$ \byref{def:V.V}. Пусть $M \magnitudeI < m \magnitudeII$, то есть $m \magnitudeII > M \magnitudeI$,\\ $\therefore M \magnitudeIII < m \magnitudeIV$, или $m \magnitudeIV > M \magnitudeIII$. @@ -9229,7 +9228,7 @@ \chapter*{Аксиомы} если $m \magnitudeII = \mbox{ или } < M \magnitudeI$,\\ то $m \magnitudeIV =, \mbox{ или } < M \magnitudeIII$. -И, следовательно \bycref{def:V.V}, мы заключаем,\\ +И, следовательно \byref{def:V.V}, мы заключаем,\\ что $\magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$. $\therefore$ если четыре величины пропорциональны… и~т. д. @@ -9257,7 +9256,7 @@ \chapter*{Аксиомы} Возьмем $M \magnitudeI$, $m \magnitudeII$, $M \magnitudeIII$, $m \magnitudeIV$. Поскольку \magnitudeI\ столько же раз кратна \magnitudeII, \\ -сколько \magnitudeIII кратна \magnitudeIV\ \bycref{\hypref},\\ +сколько \magnitudeIII кратна \magnitudeIV\ \byref{\hypref},\\ и $M \magnitudeI$ столько же раз кратна \magnitudeI, \\ столько $M \magnitudeIII$ кратна \magnitudeIII. @@ -9275,7 +9274,7 @@ \chapter*{Аксиомы} И, в~общем виде, если $M \magnitudeI >, = \mbox{ или } < m \magnitudeII$,\\ то и~$M \magnitudeIII >, = \mbox{ или } < m \magnitudeIV$. -$\therefore$ \bycref{def:V.V},\\ +$\therefore$ \byref{def:V.V},\\ $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$. Теперь пусть \magnitudeII будет такой же частью от \magnitudeI, \\ @@ -9291,7 +9290,7 @@ \chapter*{Аксиомы} Следовательно, как и~в предыдущем случае, \\ $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$. -И $\therefore \magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$ \bycref{prop:V.B}.\\ +И $\therefore \magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$ \byref{prop:V.B}.\\ $\therefore$ если первая величина… и~т. д. \end{center} @@ -9334,10 +9333,10 @@ \chapter*{Аксиомы} возьмем \magnitudeVI\ столько же раз кратную \magnitudeIV, \\ тогда, поскольку $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$\\ и мы взяли равнократные второй и~четвертой, \magnitudeI\ и~\magnitudeVI,\\ -следовательно \bycref{prop:V.IV},\\ +следовательно \byref{prop:V.IV},\\ $\magnitudeI : \magnitudeV :: \magnitudeIII : \magnitudeVI$,\\ - но \bycref{\constref} $\magnitudeI = \magnitudeV$\\ - $\therefore$ \bycref{prop:V.A} $\magnitudeIII = \magnitudeVI$\\ + но \byref{\constref} $\magnitudeI = \magnitudeV$\\ + $\therefore$ \byref{prop:V.A} $\magnitudeIII = \magnitudeVI$\\ и~\magnitudeVI\ столько же раз кратна \magnitudeIV,\\ сколько \magnitudeI\ кратна \magnitudeII. @@ -9345,7 +9344,7 @@ \chapter*{Аксиомы} а также \magnitudeII\ составляет часть \magnitudeI;\\ тогда \magnitudeIV\ будет составлять такую же часть \magnitudeIII. -Если перевернуть, $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{prop:V.B},\\ +Если перевернуть, $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{prop:V.B},\\ но \magnitudeII\ составляет часть \magnitudeI, то есть \magnitudeI\ кратна \magnitudeII. @@ -9377,13 +9376,13 @@ \chapter*{Аксиомы} $\therefore$ если $M \magnitudeI >, = \mbox{ или } < m \magnitudeIII$,\\ то $M \magnitudeII >, = \mbox{ или } < m \magnitudeIII$,\\ -и $\therefore \magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \bycref{def:V.V}. +и $\therefore \magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \byref{def:V.V}. Из вышеизложенного очевидно, что\\ если $m \magnitudeIII >, = \mbox{ или } < M \magnitudeI$,\\ то $m \magnitudeIII >, = \mbox{ или } < M \magnitudeII$. -$\therefore \magnitudeIII : \magnitudeI = \magnitudeIII : \magnitudeII$ \bycref{def:V.V}. +$\therefore \magnitudeIII : \magnitudeI = \magnitudeIII : \magnitudeII$ \byref{def:V.V}. $\therefore$ равные величины… и~т. д. \end{center} @@ -9488,7 +9487,7 @@ \chapter*{Аксиомы} то есть, $m' \magnitudeIII$ должна быть $< M' \magnitudeI$. $\therefore M' \magnitudeI > m' \magnitudeIII$, но, как было показано выше,\\ -$M' \magnitudeII \ngtr m' \magnitudeIII$, следовательно \bycref{def:V.VII},\\ \magnitudeI\ имеет к~\magnitudeIII\ большее отношение, чем $\magnitudeII : \magnitudeIII$. +$M' \magnitudeII \ngtr m' \magnitudeIII$, следовательно \byref{def:V.VII},\\ \magnitudeI\ имеет к~\magnitudeIII\ большее отношение, чем $\magnitudeII : \magnitudeIII$. Теперь докажем, что у~\magnitudeIII\ большее отношение к~меньшей величине \magnitudeII, чем к~большей \magnitudeI,\\ или $\magnitudeIII : \magnitudeII > \magnitudeIII : \magnitudeI$. @@ -9501,7 +9500,7 @@ \chapter*{Аксиомы} и $\magnitudeIII \ngtr M' \magnitudeIa$, следовательно,\\ $m' \magnitudeIII - \magnitudeIII + \magnitudeIII < M' \magnitudeIb + M' \magnitudeIa$. -$\therefore m' \magnitudeIII < M' \magnitudeI$, и~$\therefore$ \bycref{def:V.VII}, \magnitudeIII\ имеет к~\magnitudeII\ большее отношение, чем \magnitudeIII\ к~\magnitudeI. +$\therefore m' \magnitudeIII < M' \magnitudeI$, и~$\therefore$ \byref{def:V.VII}, \magnitudeIII\ имеет к~\magnitudeII\ большее отношение, чем \magnitudeIII\ к~\magnitudeI. $\therefore$ из неравных величин… и~т. д. \end{center} @@ -9520,7 +9519,7 @@ \chapter*{Аксиомы} \end{tabular} \end{center} -Следовательно, $\textcolor{byred}{64} + \textcolor{byblack}{8}$, или $72$, больше, чем $\textcolor{byblue}{70}$, но $\textcolor{byyellow}{64}$ не больше, чем $\textcolor{byblue}{70}$, $\therefore$ \bycref{def:V.VII} $\textcolor{byblack}{9}$ имеет большее отношение к~$\textcolor{byblue}{7}$, чем $\textcolor{byyellow}{8}$ к~$\textcolor{byblue}{7}$. +Следовательно, $\textcolor{byred}{64} + \textcolor{byblack}{8}$, или $72$, больше, чем $\textcolor{byblue}{70}$, но $\textcolor{byyellow}{64}$ не больше, чем $\textcolor{byblue}{70}$, $\therefore$ \byref{def:V.VII} $\textcolor{byblack}{9}$ имеет большее отношение к~$\textcolor{byblue}{7}$, чем $\textcolor{byyellow}{8}$ к~$\textcolor{byblue}{7}$. Приведенное выше — всего лишь иллюстрация к~дальнейшему доказательству, так как это свойство можно легко показать как для этих, так и~для других чисел следующим образом: поскольку, если предыдущее содержит последующее большее число раз, чем другое предыдущее содержит свое последующее, или составленная дробь с~предыдущим в~качестве числителя и~последующим в~качестве знаменателя больше другой дроби, составленной из другого предыдущего в~качестве числителя и~его последующего в~качестве знаменателя, отношение первого предыдущего к~его последующему больше, чем отношение второго предыдущего к~его последующему. @@ -9548,7 +9547,7 @@ \chapter*{Аксиомы} Пусть $\drawMagnitude{I} : \drawMagnitude{III} :: \drawMagnitude{II} : \drawMagnitude{III}$, тогда $\magnitudeI = \magnitudeII$. Поскольку если нет, пусть $\magnitudeI > \magnitudeII$,\\ -тогда $\magnitudeI : \magnitudeIII > \magnitudeII : \magnitudeIII$ \bycref{prop:V.VIII}, +тогда $\magnitudeI : \magnitudeIII > \magnitudeII : \magnitudeIII$ \byref{prop:V.VIII}, что невозможно, поскольку противоречит гипотезе.\\ $\therefore \magnitudeI \ngtr \magnitudeII$. @@ -9585,8 +9584,8 @@ \chapter*{Аксиомы} Пусть $\drawMagnitude{I} : \drawMagnitude{III} > \drawMagnitude{II} : \drawMagnitude{III}$, тогда $\magnitudeI > \magnitudeII$. Поскольку если нет, пусть $\magnitudeI = \mbox{ или } < \magnitudeII$,\\ -тогда $\magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \bycref{prop:V.VII}\\ -или $\magnitudeI : \magnitudeIII < \magnitudeII : \magnitudeIII$ \bycref{prop:V.VIII} и~(перевернув),\\ +тогда $\magnitudeI : \magnitudeIII = \magnitudeII : \magnitudeIII$ \byref{prop:V.VII}\\ +или $\magnitudeI : \magnitudeIII < \magnitudeII : \magnitudeIII$ \byref{prop:V.VIII} и~(перевернув),\\ что противоречит гипотезе. $\therefore \magnitudeI \neq \mbox{ или } < \magnitudeII$,\\ @@ -9596,8 +9595,8 @@ \chapter*{Аксиомы} тогда $\magnitudeII < \magnitudeI$. Поскольку если нет, $\magnitudeII \mbox{ должна быть } > \mbox{ или } = \magnitudeI$,\\ -тогда $\magnitudeIII : \magnitudeII < \magnitudeIII : \magnitudeI$ \bycref{prop:V.VIII} и~(перевернув),\\ -или $\magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeI$ \bycref{prop:V.VII}, что противоречит гипотезе. +тогда $\magnitudeIII : \magnitudeII < \magnitudeIII : \magnitudeI$ \byref{prop:V.VIII} и~(перевернув),\\ +или $\magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeI$ \byref{prop:V.VII}, что противоречит гипотезе. $\therefore \magnitudeII \ngtr \mbox{ или } = \magnitudeI$,\\ и $\therefore \magnitudeII \mbox{ должна быть } < \magnitudeI$. @@ -9625,10 +9624,10 @@ \chapter*{Аксиомы} Поскольку если $M \magnitudeI >, = \mbox{ или } < m \magnitudeII$,\\ то $M \magnitudeV >, = \mbox{ или } < m \magnitudeVI$,\\ и если $M \magnitudeV >, = \mbox{ или } < m \magnitudeVI$,\\ -то $M \magnitudeIII >, = \mbox{ или } < m \magnitudeIV$ \bycref{def:V.V}. +то $M \magnitudeIII >, = \mbox{ или } < m \magnitudeIV$ \byref{def:V.V}. $\therefore$ if $M \magnitudeI >, = \mbox{ или } < m \magnitudeII$, $M \magnitudeIII >, = \mbox{ или } < m \magnitudeIV$,\\ -и $\therefore$ \bycref{def:V.V}, $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$. +и $\therefore$ \byref{def:V.V}, $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$. $\therefore$ отношения, тождественные между собой… и~т. д. \end{center} @@ -9662,13 +9661,13 @@ \chapter*{Аксиомы} Поскольку если $M \magnitudeI > m \magnitudeII$, то $M \magnitudeIII > m \magnitudeIV$,\\ и $M \magnitudeV > m \magnitudeVI$, $M \magnitudeVII > m \magnitudeVIII$,\\ -а также $M \magnitudeIX > m \magnitudeX$ \bycref{def:V.V}. +а также $M \magnitudeIX > m \magnitudeX$ \byref{def:V.V}. Следовательно, если $M \magnitudeI > m \magnitudeII$,\\ то $M \magnitudeI + M \magnitudeIII + M \magnitudeV + M \magnitudeVII + M \magnitudeIX$, или $M (\magnitudeI + \magnitudeIII + \magnitudeV + \magnitudeVII + \magnitudeIX)$ будет больше, чем $m \magnitudeII + m \magnitudeIV + m \magnitudeVI + m \magnitudeVIII + m \magnitudeX$, или $m (\magnitudeII + \magnitudeIV + \magnitudeVI + \magnitudeVIII + \magnitudeX)$. \end{center} -Так же можно показать, что если $M$ раз одно из предыдущих равно или меньше, чем $m$ раз одно из последующих, $M$ раз все предыдущие будут равны или меньше, чем $m$ раз все последующие вместе. Следовательно \bycref{def:V.V}, как одно из предыдущих к~последующему, так и~все предыдущие ко всем последующим вместе. +Так же можно показать, что если $M$ раз одно из предыдущих равно или меньше, чем $m$ раз одно из последующих, $M$ раз все предыдущие будут равны или меньше, чем $m$ раз все последующие вместе. Следовательно \byref{def:V.V}, как одно из предыдущих к~последующему, так и~все предыдущие ко всем последующим вместе. $\therefore$ если несколько величин… и~т. д. @@ -9692,18 +9691,18 @@ \chapter*{Аксиомы} Поскольку $\magnitudeIII : \magnitudeIV > \magnitudeV : \magnitudeVI$, есть такие кратные ($M'$ и~$m'$) величин \magnitudeIII\ и~\magnitudeV и~величин \magnitudeIV\ и~\magnitudeVI,\\ что $M' \magnitudeIII > m' \magnitudeIV$,\\ -но $M' \magnitudeV \ngtr m' \magnitudeVI$ \bycref{def:V.VII}. +но $M' \magnitudeV \ngtr m' \magnitudeVI$ \byref{def:V.VII}. Возьмем эти кратные и~возьмем такие же кратные \magnitudeI\ и~\magnitudeII. -$\therefore$ \bycref{def:V.V}, если $M' \magnitudeI >, =, \mbox{ или } < m' \magnitudeII$,\\ +$\therefore$ \byref{def:V.V}, если $M' \magnitudeI >, =, \mbox{ или } < m' \magnitudeII$,\\ то $M' \magnitudeIII >, =, \mbox{ и~} < m' \magnitudeIV$,\\ -но $M' \magnitudeIII > m' \magnitudeIV$ \bycref{\constref}; +но $M' \magnitudeIII > m' \magnitudeIV$ \byref{\constref}; $\therefore M' \magnitudeI > m' \magnitudeII$,\\ -но $M' \magnitudeV \ngtr m' \magnitudeVI$ \bycref{\constref} +но $M' \magnitudeV \ngtr m' \magnitudeVI$ \byref{\constref} -И, следовательно, \bycref{def:V.VII},\\ +И, следовательно, \byref{def:V.VII},\\ $\magnitudeI : \magnitudeII > \magnitudeV : \magnitudeVI$. $\therefore$ если первая величина ко второй… и~т. д. @@ -9724,19 +9723,19 @@ \chapter*{Аксиомы} Пусть $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$,\\ и предположим, что $\magnitudeI > \magnitudeIII$, тогда $\magnitudeII > \magnitudeIV$. -Поскольку $\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \bycref{prop:V.VIII}\\ +Поскольку $\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \byref{prop:V.VIII}\\ и, согласно гипотезе, $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$. -$\therefore \magnitudeIII : \magnitudeIV > \magnitudeIII : \magnitudeII$ \bycref{prop:V.XIII}. +$\therefore \magnitudeIII : \magnitudeIV > \magnitudeIII : \magnitudeII$ \byref{prop:V.XIII}. -$\therefore \magnitudeIV < \magnitudeII$ \bycref{prop:V.X} или $\magnitudeII > \magnitudeIV$. +$\therefore \magnitudeIV < \magnitudeII$ \byref{prop:V.X} или $\magnitudeII > \magnitudeIV$. Теперь пусть $\magnitudeI = \magnitudeIII$, тогда $\magnitudeII = \magnitudeIV$. -Поскольку $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \bycref{prop:V.VII}\\ -и $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$ \bycref{\hypref};\\ -$\therefore \magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XI}\\ -и $\therefore \magnitudeII = \magnitudeIV$ \bycref{prop:V.IX}. +Поскольку $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \byref{prop:V.VII}\\ +и $\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeIV$ \byref{\hypref};\\ +$\therefore \magnitudeIII : \magnitudeII = \magnitudeIII : \magnitudeIV$ \byref{prop:V.XI}\\ +и $\therefore \magnitudeII = \magnitudeIV$ \byref{prop:V.IX}. Теперь же, если $\magnitudeI < \magnitudeIII$, то $\magnitudeII < \magnitudeIV$,\\ поскольку $\magnitudeIII > \magnitudeI$ и~$\magnitudeIII : \magnitudeIV = \magnitudeI : \magnitudeII$. @@ -9766,7 +9765,7 @@ \chapter*{Аксиомы} &= \magnitudeI : \magnitudeII \mbox{,} \\ \end{aligned}$ -$\therefore \magnitudeI : \magnitudeII :: 4 \magnitudeI : 4 \magnitudeII$ \bycref{prop:V.XII}. +$\therefore \magnitudeI : \magnitudeII :: 4 \magnitudeI : 4 \magnitudeII$ \byref{prop:V.XII}. И, поскольку те же рассуждения применимы в~общем, получим:\\ $\magnitudeI : \magnitudeII :: M' \magnitudeI : M' \magnitudeII$. @@ -9808,15 +9807,15 @@ \chapter*{Аксиомы} Пусть $\drawMagnitude{I} : \drawMagnitude{II} :: \drawMagnitude{III} : \drawMagnitude{IV}$,\\ тогда $\magnitudeI : \magnitudeIII :: \magnitudeII : \magnitudeIV$. -Поскольку $M \magnitudeI : M \magnitudeII :: \magnitudeI : \magnitudeII$ \bycref{prop:V.XV},\\ -и $M \magnitudeI : M \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{\hypref,prop:V.XI},\\ -а также $m \magnitudeIII : m \magnitudeIV :: \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XV}. +Поскольку $M \magnitudeI : M \magnitudeII :: \magnitudeI : \magnitudeII$ \byref{prop:V.XV},\\ +и $M \magnitudeI : M \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{\hypref,prop:V.XI},\\ +а также $m \magnitudeIII : m \magnitudeIV :: \magnitudeIII : \magnitudeIV$ \byref{prop:V.XV}. -$\therefore M \magnitudeI : M \magnitudeII :: m \magnitudeIII : m \magnitudeIV$ \bycref{prop:V.XIV},\\ +$\therefore M \magnitudeI : M \magnitudeII :: m \magnitudeIII : m \magnitudeIV$ \byref{prop:V.XIV},\\ и $\therefore$ если $M \magnitudeI >, = \mbox { или } < m \magnitudeIII$,\\ -то $M \magnitudeII >, =, \mbox{ или } < m \magnitudeIV$ \bycref{prop:V.XIV}. +то $M \magnitudeII >, =, \mbox{ или } < m \magnitudeIV$ \byref{prop:V.XIV}. -следовательно \bycref{def:V.V},\\ +следовательно \byref{def:V.V},\\ $\magnitudeI : \magnitudeIII :: \magnitudeII : \magnitudeIV$. $\therefore$ если четыре величины… и~т. д. @@ -9863,9 +9862,9 @@ \chapter*{Аксиомы} Возьмем $\magnitudeI > m \magnitudeII$, к~каждой добавим $M \magnitudeII$,\\ тогда получим $M \magnitudeI + M \magnitudeII > m \magnitudeII + M \magnitudeII$,\\ или $M (\magnitudeI + \magnitudeII) > (m + M) \magnitudeII$,\\ -но, поскольку $\magnitudeI + \magnitudeII: \magnitudeII :: \magnitudeIII + \magnitudeIV: \magnitudeIV$ \bycref{\hypref},\\ +но, поскольку $\magnitudeI + \magnitudeII: \magnitudeII :: \magnitudeIII + \magnitudeIV: \magnitudeIV$ \byref{\hypref},\\ и $M (\magnitudeI + \magnitudeII) > (m + M) \magnitudeII$,\\ -$\therefore M(\magnitudeIII + \magnitudeIV) > (m + M) \magnitudeIV$ \bycref{def:V.V}. +$\therefore M(\magnitudeIII + \magnitudeIV) > (m + M) \magnitudeIV$ \byref{def:V.V}. $\therefore M \magnitudeIII + M \magnitudeIV > m \magnitudeIV + M \magnitudeIV$. @@ -9873,7 +9872,7 @@ \chapter*{Аксиомы} то есть когда $M \magnitudeI > m \magnitudeII$, тогда $M \magnitudeIII > m \magnitudeIV$. Так же можно доказать, что если $M \magnitudeI = \mbox{ или } < m \magnitudeII$, то $M \magnitudeIII = \mbox{ или } < m \magnitudeIV$,\\ -и $\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{def:V.V}. +и $\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{def:V.V}. $\therefore$ если величины пропорциональны составленные… и~т. д. \end{center} @@ -9922,12 +9921,12 @@ \chapter*{Аксиомы} Ведь если нет, то $\magnitudeI + \magnitudeII : \magnitudeII :: \magnitudeIII + \drawMagnitude{V} : \magnitudeV$,\\ полагая, что $\magnitudeV \neq \magnitudeIV$. -$\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeV$ \bycref{prop:V.XVII},\\ -но $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \bycref{\hypref}. +$\therefore \magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeV$ \byref{prop:V.XVII},\\ +но $\magnitudeI : \magnitudeII :: \magnitudeIII : \magnitudeIV$ \byref{\hypref}. -$\therefore \magnitudeIII : \magnitudeV :: \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XI}. +$\therefore \magnitudeIII : \magnitudeV :: \magnitudeIII : \magnitudeIV$ \byref{prop:V.XI}. -$\therefore \magnitudeV = \magnitudeIV$ \bycref{prop:V.IX},\\ +$\therefore \magnitudeV = \magnitudeIV$ \byref{prop:V.IX},\\ что противоречит предположению. $\therefore \magnitudeV \mbox{ не неравна } \magnitudeIV$;\\ @@ -9959,9 +9958,9 @@ \chapter*{Аксиомы} Теперь $\magnitudeII : \magnitudeIV :: \magnitudeI : \magnitudeIII$ (перестав.). -Но $\magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV :: \magnitudeI : \magnitudeIII$ \bycref{\hypref}. +Но $\magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV :: \magnitudeI : \magnitudeIII$ \byref{\hypref}. -$\therefore \magnitudeII : \magnitudeIV :: \magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV$ \bycref{prop:V.XI}. +$\therefore \magnitudeII : \magnitudeIV :: \magnitudeI + \magnitudeII : \magnitudeIII + \magnitudeIV$ \byref{prop:V.XI}. $\therefore$ если целая величина относится к~целой… и~т. д. \end{center} @@ -10040,7 +10039,7 @@ \chapter*{Аксиомы} \def\varO{\textcolor{byblue}{O}} \def\varP{\textcolor{byred}{P}} \def\varQ{\textcolor{byred}{Q}} -\enquote{По равенству в~перемешанной пропорции} говорят, когда первая величина ко второй в~первом ряду относится, как последняя к~предпоследней во втором ряду, вторая к~третьей в~первом ряду, как пред-предпоследняя к~предпоследней во втором, третья к~четвертой в~первом, как четвертая с~конца к~пред-предпоследней во втором и~так далее, и~утверждается то же, что и~в \bycref{def:V.XVIII}. Это иллюстрируется в~\bycref{prop:V.XXIII}. +\enquote{По равенству в~перемешанной пропорции} говорят, когда первая величина ко второй в~первом ряду относится, как последняя к~предпоследней во втором ряду, вторая к~третьей в~первом ряду, как пред-предпоследняя к~предпоследней во втором, третья к~четвертой в~первом, как четвертая с~конца к~пред-предпоследней во втором и~так далее, и~утверждается то же, что и~в \byref{def:V.XVIII}. Это иллюстрируется в~\byref{prop:V.XXIII}. \begin{center} Так, если есть два ряда величин,\\ @@ -10077,10 +10076,10 @@ \chapter*{Аксиомы} $\magnitudeI : \magnitudeIV :: \magnitudeII : \magnitudeV$\\ и $\magnitudeII : \magnitudeV :: \magnitudeIII : \magnitudeVI$. -$\therefore \magnitudeI : \magnitudeIV :: \magnitudeIII : \magnitudeVI$ \bycref{prop:V.XI}. +$\therefore \magnitudeI : \magnitudeIV :: \magnitudeIII : \magnitudeVI$ \byref{prop:V.XI}. $\therefore$ если $\magnitudeI >, =, \mbox{ или } < \magnitudeIII$,\\ -тогда $\magnitudeIV >, =, \mbox{ или } < \magnitudeVI$ \bycref{prop:V.XIV}. +тогда $\magnitudeIV >, =, \mbox{ или } < \magnitudeVI$ \byref{prop:V.XIV}. $\therefore$ если есть три величины… и~т. д. \end{center} @@ -10109,28 +10108,28 @@ \chapter*{Аксиомы} Допустим, $\magnitudeI > \magnitudeIII$,\\ тогда, поскольку \magnitudeII\ любая другая величина,\\ -$\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \bycref{prop:V.VIII}. +$\magnitudeI : \magnitudeII > \magnitudeIII : \magnitudeII$ \byref{prop:V.VIII}. -Но $\magnitudeV : \magnitudeVI :: \magnitudeI : \magnitudeII$ \bycref{\hypref}. +Но $\magnitudeV : \magnitudeVI :: \magnitudeI : \magnitudeII$ \byref{\hypref}. -$\therefore \magnitudeV : \magnitudeVI > \magnitudeIII : \magnitudeII$ \bycref{prop:V.XIII},\\ -и, поскольку $\magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \bycref{\hypref},\\ +$\therefore \magnitudeV : \magnitudeVI > \magnitudeIII : \magnitudeII$ \byref{prop:V.XIII},\\ +и, поскольку $\magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \byref{\hypref},\\ $\therefore \magnitudeIII : \magnitudeII :: \magnitudeV : \magnitudeIV$ (перев.),\\ и было показано, что $\magnitudeV : \magnitudeVI > \magnitudeIII : \magnitudeII$, -$\therefore \magnitudeV : \magnitudeVI > \magnitudeV : \magnitudeIV$ \bycref{prop:V.XIII}. +$\therefore \magnitudeV : \magnitudeVI > \magnitudeV : \magnitudeIV$ \byref{prop:V.XIII}. $\therefore \magnitudeVI < \magnitudeIV$, то есть $\magnitudeIV > \magnitudeVI$. Теперь, пусть $\magnitudeI = \magnitudeIII$; тогда $\magnitudeIV = \magnitudeVI$.\\ Поскольку $\magnitudeI = \magnitudeIII$,\\ -$\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \bycref{prop:V.VII}. +$\magnitudeI : \magnitudeII = \magnitudeIII : \magnitudeII$ \byref{prop:V.VII}. -Но $\magnitudeI : \magnitudeII = \magnitudeV : \magnitudeVI$ \bycref{\hypref}\\ +Но $\magnitudeI : \magnitudeII = \magnitudeV : \magnitudeVI$ \byref{\hypref}\\ и $\magnitudeIII : \magnitudeII = \magnitudeV : \magnitudeIV$ (\hypstr\ и~перев.). -$\therefore \magnitudeV : \magnitudeVI = \magnitudeV : \magnitudeIV$ \bycref{prop:V.XI}. +$\therefore \magnitudeV : \magnitudeVI = \magnitudeV : \magnitudeIV$ \byref{prop:V.XI}. -$\therefore \magnitudeIV = \magnitudeVI$ \bycref{prop:V.IX}. +$\therefore \magnitudeIV = \magnitudeVI$ \byref{prop:V.IX}. И теперь пусть $\magnitudeI < \magnitudeIII$, тогда $\magnitudeIV < \magnitudeVI$;\\ поскольку $\magnitudeIII > \magnitudeI$,\\ @@ -10179,7 +10178,7 @@ \chapter*{Аксиомы} $\magnitudeI, \magnitudeII, \magnitudeIII, \magnitudeIV, \magnitudeV, \magnitudeVI$\\ и $M \magnitudeI, m \magnitudeII, N \magnitudeIII, M \magnitudeIV, m \magnitudeV, N \magnitudeVI,$\\ поскольку $\magnitudeI : \magnitudeII :: \magnitudeIV : \magnitudeV$,\\ -$\therefore M \magnitudeI : m \magnitudeII :: M \magnitudeIV : m \magnitudeV$ \bycref{prop:V.IV}. +$\therefore M \magnitudeI : m \magnitudeII :: M \magnitudeIV : m \magnitudeV$ \byref{prop:V.IV}. По той же причине\\ $m \magnitudeII : N \magnitudeIII :: m \magnitudeV : N \magnitudeVI$;\\ @@ -10189,8 +10188,8 @@ \chapter*{Аксиомы} которые, взятые по две из каждого ряда, имеют равные отношения. $\therefore$ если $M \magnitudeI >, = \mbox{ или } < N \magnitudeIII$,\\ -то $M \magnitudeIV >, = \mbox{ или } < N \magnitudeVI$ \bycref{prop:V.XX},\\ -и $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeVI$ \bycref{def:V.V}. +то $M \magnitudeIV >, = \mbox{ или } < N \magnitudeVI$ \byref{prop:V.XX},\\ +и $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeVI$ \byref{def:V.V}. Теперь, допустим, есть четыре величины \drawMagnitude{Ia}, \drawMagnitude{IIa}, \drawMagnitude{IIIa}, \drawMagnitude{IVa}\\ и другие четыре \drawMagnitude{Va}, \drawMagnitude{VIa}, \drawMagnitude{VIIa}, \drawMagnitude{VIIIa},\\ @@ -10242,22 +10241,22 @@ \chapter*{Аксиомы} \begin{center} $\magnitudeI, \magnitudeII, \magnitudeIII, \magnitudeIV, \magnitudeV, \magnitudeVI,$\\ $M \magnitudeI, M \magnitudeII, m \magnitudeIII, M \magnitudeIV, m \magnitudeV, m \magnitudeVI,$\\ -тогда $\magnitudeI : \magnitudeII :: M \magnitudeI : M \magnitudeII$ \bycref{prop:V.XV};\\ +тогда $\magnitudeI : \magnitudeII :: M \magnitudeI : M \magnitudeII$ \byref{prop:V.XV};\\ и по той же причине\\ $\magnitudeV : \magnitudeVI :: m \magnitudeV : m \magnitudeVI$;\\ -но $\magnitudeI : \magnitudeII :: \magnitudeV : \magnitudeVI$ \bycref{\hypref},\\ -$\therefore M \magnitudeI : M \magnitudeII :: \magnitudeV : \magnitudeVI$ \bycref{prop:V.XI}. +но $\magnitudeI : \magnitudeII :: \magnitudeV : \magnitudeVI$ \byref{\hypref},\\ +$\therefore M \magnitudeI : M \magnitudeII :: \magnitudeV : \magnitudeVI$ \byref{prop:V.XI}. -И поскольку $\magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \bycref{\hypref},\\ -$\therefore M \magnitudeII : m \magnitudeIII :: M \magnitudeIV : m \magnitudeV$ \bycref{prop:V.IV}. +И поскольку $\magnitudeII : \magnitudeIII :: \magnitudeIV : \magnitudeV$ \byref{\hypref},\\ +$\therefore M \magnitudeII : m \magnitudeIII :: M \magnitudeIV : m \magnitudeV$ \byref{prop:V.IV}. Тогда, поскольку есть три величины\\ $M \magnitudeI, M \magnitudeII, m \magnitudeII$\\ и другие три $M \magnitudeIV, m \magnitudeV, m \magnitudeVI$,\\ которые, взятые по две в~противоположном порядке, имеют равные отношения,\\ значит, если $M \magnitudeI >, =, \mbox{ или } < m \magnitudeIII$,\\ -то $M \magnitudeIV >, =, \mbox{ или } < m \magnitudeVI$ \bycref{prop:V.XXI},\\ -и $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeIV$ \bycref{def:V.V}. +то $M \magnitudeIV >, =, \mbox{ или } < m \magnitudeVI$ \byref{prop:V.XXI},\\ +и $\therefore \magnitudeI : \magnitudeIII :: \magnitudeIV : \magnitudeIV$ \byref{def:V.V}. Теперь пусть будет четыре величины,\\ \magnitudeI, \magnitudeII, \magnitudeIII, \magnitudeIV,\\ @@ -10305,16 +10304,16 @@ \chapter*{Аксиомы} и $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$,\\ то $\magnitudeI + \magnitudeV : \magnitudeII :: \magnitudeIII + \magnitudeVI : \magnitudeIV$. -Поскольку $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \bycref{\hypref}\\ +Поскольку $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \byref{\hypref}\\ и $\magnitudeII : \magnitudeI :: \magnitudeIV : \magnitudeIII$ (\hypstr и~перев.). -$\therefore \magnitudeV : \magnitudeI :: \magnitudeVI : \magnitudeIII$ \bycref{prop:V.XXII};\\ +$\therefore \magnitudeV : \magnitudeI :: \magnitudeVI : \magnitudeIII$ \byref{prop:V.XXII};\\ и, поскольку величины пропорциональны, они будут пропорциональны и~если их присоединить. -$\therefore \magnitudeI + \magnitudeV : \magnitudeV :: \magnitudeVI + \magnitudeIII : \magnitudeVI$ \bycref{prop:V.XVIII},\\ -но $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \bycref{\hypref}. +$\therefore \magnitudeI + \magnitudeV : \magnitudeV :: \magnitudeVI + \magnitudeIII : \magnitudeVI$ \byref{prop:V.XVIII},\\ +но $\magnitudeV : \magnitudeII :: \magnitudeVI : \magnitudeIV$ \byref{\hypref}. -$\therefore \magnitudeI + \magnitudeV : \magnitudeII :: \magnitudeVI + \magnitudeIII : \magnitudeIV$ \bycref{prop:V.XXII}. +$\therefore \magnitudeI + \magnitudeV : \magnitudeII :: \magnitudeVI + \magnitudeIII : \magnitudeIV$ \byref{prop:V.XXII}. $\therefore$ если первая величина ко второй… и~т. д. \end{center} @@ -10334,15 +10333,15 @@ \chapter*{Аксиомы} Пусть четыре величины $\drawMagnitude{Ia} + \drawMagnitude{III}, \drawMagnitude{IIa} + \drawMagnitude{IV}, \magnitudeIII \mbox{ и~} \magnitudeIV$ будут пропорциональными. $\magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV :: \magnitudeIII : \magnitudeIV$,\\ -и пусть $\magnitudeIa + \magnitudeIII$ будет наибольшей из четырех, следовательно, согласно \bycref{prop:V.A} и~\bycref{prop:V.XIV}, \magnitudeIV\ будет наименьшей. +и пусть $\magnitudeIa + \magnitudeIII$ будет наибольшей из четырех, следовательно, согласно \byref{prop:V.A} и~\byref{prop:V.XIV}, \magnitudeIV\ будет наименьшей. Тогда $\magnitudeIa + \magnitudeIII + \magnitudeIV > \magnitudeIIa + \magnitudeIV + \magnitudeIII$,\\ поскольку $\magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV :: \magnitudeIII : \magnitudeIV$. -$\therefore \magnitudeIa : \magnitudeIIa :: \magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV$ \bycref{prop:V.XIX},\\ -но $\magnitudeIa + \magnitudeIII > \magnitudeIIa + \magnitudeIV$ \bycref{\hypref}. +$\therefore \magnitudeIa : \magnitudeIIa :: \magnitudeIa + \magnitudeIII : \magnitudeIIa + \magnitudeIV$ \byref{prop:V.XIX},\\ +но $\magnitudeIa + \magnitudeIII > \magnitudeIIa + \magnitudeIV$ \byref{\hypref}. -$\therefore \magnitudeIa > \magnitudeIIa$ \bycref{prop:V.A}. +$\therefore \magnitudeIa > \magnitudeIIa$ \byref{prop:V.A}. К каждой добавим $\magnitudeIII + \magnitudeIV$,\\ $\therefore \magnitudeIa + \magnitudeIII + \magnitudeIV > \magnitudeIIa + \magnitudeIII + \magnitudeIV$. @@ -10781,7 +10780,7 @@ \chapter*{Аксиомы} =$\\ $\dfrac {h \times k \times l \times m \times n} -{k \times l \times m \times n \times p}$ \bycref{\hypref}. +{k \times l \times m \times n \times p}$ \byref{\hypref}. Или $\dfrac{\vara \times \varb}{\varb \times \varc} \times @@ -10822,7 +10821,7 @@ \chapter*{Аксиомы} =$\\ $\dfrac {h \times k \times l \times m \times n} -{k \times l \times m \times n \times p}$ \bycref{\hypref}. +{k \times l \times m \times n \times p}$ \byref{\hypref}. И $\dfrac {m \times n} @@ -10830,7 +10829,7 @@ \chapter*{Аксиомы} = \dfrac {e \times f} -{f \times g}$ \bycref{\hypref}, +{f \times g}$ \byref{\hypref}, $\therefore \dfrac {h \times k \times l \times m \times n} @@ -10986,7 +10985,7 @@ \chapter*{Определения} draw byNamedLine(MB,BC); draw byLabelsOnPolygon(A, C, M)(ALL_LABELS, 0); }, -образованные таким образом, равны между собой, поскольку равны их основания \bycref{prop:I.XXXVIII}. +образованные таким образом, равны между собой, поскольку равны их основания \byref{prop:I.XXXVIII}. $\therefore$ \trianglesAMC\ и~его основание соответственно равнократны \polygonABC\ и~его основанию \drawUnitLine{BC}. @@ -10998,13 +10997,13 @@ \chapter*{Определения} } и его основание соответственно равнократны \polygonACD\ и~его основанию \drawUnitLine{CD}. -$\therefore$ если $m$ или $6$ раз \polygonABC\ $>, = \mbox{ или } < n$ или $5$ раз \polygonACD, то $m$ или $6$ раз \drawUnitLine{BC} $>, = \mbox{ или } < n$ или $5$ раз \drawUnitLine{CD}, $m$ и~$n$ обозначают любые кратные, взятые как в~\bycref{def:V.V}. Хотя мы показали только, что это свойство проявляется, когда $m = 6$ и~$n = 5$, но очевидно, что это свойство работает для любых кратных значений, которые можно придать $m$ и~$n$. +$\therefore$ если $m$ или $6$ раз \polygonABC\ $>, = \mbox{ или } < n$ или $5$ раз \polygonACD, то $m$ или $6$ раз \drawUnitLine{BC} $>, = \mbox{ или } < n$ или $5$ раз \drawUnitLine{CD}, $m$ и~$n$ обозначают любые кратные, взятые как в~\byref{def:V.V}. Хотя мы показали только, что это свойство проявляется, когда $m = 6$ и~$n = 5$, но очевидно, что это свойство работает для любых кратных значений, которые можно придать $m$ и~$n$. \begin{center} -$\therefore \polygonABC\ : \polygonACD\ :: \drawUnitLine{BC} : \drawUnitLine{CD}$ \bycref{def:V.V}. +$\therefore \polygonABC\ : \polygonACD\ :: \drawUnitLine{BC} : \drawUnitLine{CD}$ \byref{def:V.V}. \end{center} -Параллелограммы с~той же высотой вдвое больше треугольников на тех же основаниях и~пропорциональны им (часть I), и, поскольку они вдвое больше, такие параллелограммы относятся, как их основания \bycref{prop:V.XV}. +Параллелограммы с~той же высотой вдвое больше треугольников на тех же основаниях и~пропорциональны им (часть I), и, поскольку они вдвое больше, такие параллелограммы относятся, как их основания \byref{prop:V.XV}. \qed @@ -11082,13 +11081,13 @@ \chapter*{Определения} тогда $\drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$. Проведем \drawUnitLine{BE} и~\drawUnitLine{CD},\\ -и $\drawLine[middle][triangleBDE]{DE,BE,DB} = \drawLine[middle][triangleCDE]{EC,CD,DE}$ \bycref{prop:I.XXXVII}. +и $\drawLine[middle][triangleBDE]{DE,BE,DB} = \drawLine[middle][triangleCDE]{EC,CD,DE}$ \byref{prop:I.XXXVII}. $\therefore \triangleBDE\ : -\drawLine[middle][triangleADE]{AD,AE,DE} :: \triangleCDE\ : \triangleADE$ \bycref{prop:V.VII}. +\drawLine[middle][triangleADE]{AD,AE,DE} :: \triangleCDE\ : \triangleADE$ \byref{prop:V.VII}. -Но $\triangleBDE\ : \triangleADE\ :: \drawUnitLine{DB} : \drawUnitLine{AD}$ \bycref{prop:VI.I},\\ -$\therefore \drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \bycref{prop:V.XI}. +Но $\triangleBDE\ : \triangleADE\ :: \drawUnitLine{DB} : \drawUnitLine{AD}$ \byref{prop:VI.I},\\ +$\therefore \drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \byref{prop:V.XI}. \end{center} \vfill\pagebreak @@ -11104,16 +11103,16 @@ \chapter*{Определения} \mbox{ поскольку } \drawUnitLine{DB} : \drawUnitLine{AD} &:: \triangleBDE\ : \triangleADE \\ \mbox{ и~} \drawUnitLine{EC} : \drawUnitLine{AE} &:: \triangleCDE\ : \triangleADE \\ \end{aligned} -\right\}\mbox{ \bycref{prop:VI.I} }$. +\right\}\mbox{ \byref{prop:VI.I} }$. -Но $\drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \bycref{\hypref}. +Но $\drawUnitLine{DB} : \drawUnitLine{AD} :: \drawUnitLine{EC} : \drawUnitLine{AE}$ \byref{\hypref}. -$\therefore \triangleBDE\ : \triangleADE :: \triangleCDE\ : \triangleADE$ \bycref{prop:V.XI}. +$\therefore \triangleBDE\ : \triangleADE :: \triangleCDE\ : \triangleADE$ \byref{prop:V.XI}. -$\therefore \triangleBDE\ = \triangleCDE$ \bycref{prop:V.IX}. +$\therefore \triangleBDE\ = \triangleCDE$ \byref{prop:V.IX}. Но они на одном основании \drawUnitLine{BC} и~по одну его сторону,\\ -и $\therefore \drawUnitLine{DE} \parallel \drawUnitLine{BC}$ \bycref{prop:I.XXXIX}. +и $\therefore \drawUnitLine{DE} \parallel \drawUnitLine{BC}$ \byref{prop:I.XXXIX}. \end{center} \qed @@ -11151,18 +11150,18 @@ \chapter*{Определения} \startsubproposition{Часть I.} \begin{center} Проведем $\drawUnitLine{CE} \parallel \drawUnitLine{AD}$, до \drawUnitLine{AE},\\ -тогда $\drawAngle{BAD} = \drawAngle{E}$ \bycref{prop:I.XXIX}. +тогда $\drawAngle{BAD} = \drawAngle{E}$ \byref{prop:I.XXIX}. $\therefore \drawAngle{DAC} = \drawAngle{E}$; но $\drawAngle{DAC} = \drawAngle{C}$. $\therefore \drawAngle{C} = \drawAngle{E}$. -$\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \bycref{prop:I.VI},\\ +$\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \byref{prop:I.VI},\\ и поскольку $\drawUnitLine{AD} \parallel \drawUnitLine{CE}$,\\ -$\drawUnitLine{AE} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \bycref{prop:VI.II}. +$\drawUnitLine{AE} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \byref{prop:VI.II}. Но $\drawUnitLine{AE} = \drawUnitLine{CA}$;\\ -$\therefore \drawUnitLine{CA} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \bycref{prop:V.VII}. +$\therefore \drawUnitLine{CA} : \drawUnitLine{AB} :: \drawUnitLine{DC} : \drawUnitLine{BD}$ \byref{prop:V.VII}. \end{center} \vfill\pagebreak @@ -11170,17 +11169,17 @@ \chapter*{Определения} \startsubproposition{Часть II.} \begin{center} Оставим то же построение,\\ -и $\drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{BD} : \drawUnitLine{DC}$ \bycref{prop:VI.II}. +и $\drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{BD} : \drawUnitLine{DC}$ \byref{prop:VI.II}. -Но $\drawUnitLine{BD} : \drawUnitLine{DC} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \bycref{\hypref}. +Но $\drawUnitLine{BD} : \drawUnitLine{DC} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \byref{\hypref}. -$\therefore \drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \bycref{prop:V.XI}. +$\therefore \drawUnitLine{AB} : \drawUnitLine{AE} :: \drawUnitLine{AB} : \drawUnitLine{CA}$ \byref{prop:V.XI}. -И $\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \bycref{prop:V.IX},\\ -и $\therefore \drawAngle{E} = \drawAngle{C}$ \bycref{prop:I.V}. +И $\therefore \drawUnitLine{AE} = \drawUnitLine{CA}$ \byref{prop:V.IX},\\ +и $\therefore \drawAngle{E} = \drawAngle{C}$ \byref{prop:I.V}. Но поскольку $\drawUnitLine{AD} \parallel \drawUnitLine{CE}$, $\drawAngle{DAC} = \drawAngle{C}$,\\ -и $\drawAngle{BAD} = \drawAngle{E}$ \bycref{prop:I.XXIX},\\ +и $\drawAngle{BAD} = \drawAngle{E}$ \byref{prop:I.XXIX},\\ $\therefore \drawAngle{C} = \drawAngle{E}$, и~$\drawAngle{BAD} = \drawAngle{DAC}$,\\ и $\therefore$ \drawUnitLine{AD} делит \drawAngle{BAD,DAC} пополам. \end{center} @@ -11222,7 +11221,7 @@ \chapter*{Определения} \begin{center} Проведем \drawUnitLine{AF} и~\drawUnitLine{DF}. -Тогда, поскольку $\drawAngle{BCA} = \drawAngle{E}$, $\drawUnitLine{CA} \parallel \drawUnitLine{DF,DE}$ \bycref{prop:I.XXVIII}. +Тогда, поскольку $\drawAngle{BCA} = \drawAngle{E}$, $\drawUnitLine{CA} \parallel \drawUnitLine{DF,DE}$ \byref{prop:I.XXVIII}. И по той же причине $\drawUnitLine{CD} \parallel \drawUnitLine{AB,AF}$. @@ -11236,11 +11235,11 @@ \chapter*{Определения} } параллелограмм. -Но $\drawUnitLine{BC} : \drawUnitLine{EC} :: \drawUnitLine{DF} : \drawUnitLine{DE}$ \bycref{prop:VI.II},\\ -и поскольку $\drawUnitLine{DF} = \drawUnitLine{CA}$ \bycref{prop:I.XXXIV},\\ +Но $\drawUnitLine{BC} : \drawUnitLine{EC} :: \drawUnitLine{DF} : \drawUnitLine{DE}$ \byref{prop:VI.II},\\ +и поскольку $\drawUnitLine{DF} = \drawUnitLine{CA}$ \byref{prop:I.XXXIV},\\ $\drawUnitLine{BC} : \drawUnitLine{EC} :: \drawUnitLine{CA} : \drawUnitLine{DE}$. -И, переставлением, $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EC} : \drawUnitLine{DE}$ \bycref{prop:V.XVI}. +И, переставлением, $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EC} : \drawUnitLine{DE}$ \byref{prop:V.XVI}. Так же можно показать, что\\ $\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{BC} : \drawUnitLine{EC}$,\\ @@ -11249,7 +11248,7 @@ \chapter*{Определения} но уже было доказано, что\\ $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EC} : \drawUnitLine{DE}$,\\ и значит по равенству\\ -$\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{CD} : \drawUnitLine{DE}$ \bycref{prop:V.XXII}. +$\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{CD} : \drawUnitLine{DE}$ \byref{prop:V.XXII}. \end{center} Следовательно, стороны при равных углах пропорциональны, а~противолежащие им соответственны. @@ -11301,24 +11300,24 @@ \chapter*{Определения} } \begin{center} Из концов \drawUnitLine{EF} проведем \drawUnitLine{FG} и~\drawUnitLine{GE},\\ -делая $\drawAngle{GEF} = \drawAngle{B}$, $\drawAngle{EFG} = \drawAngle{C}$ \bycref{prop:I.XXIII},\\ -получим $\drawAngle{G} = \drawAngle{A}$ \bycref{prop:I.XXXII}. +делая $\drawAngle{GEF} = \drawAngle{B}$, $\drawAngle{EFG} = \drawAngle{C}$ \byref{prop:I.XXIII},\\ +получим $\drawAngle{G} = \drawAngle{A}$ \byref{prop:I.XXXII}. И поскольку треугольники равноугольные,\\ -$\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{GE} : \drawUnitLine{EF}$ \bycref{prop:VI.IV}. +$\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{GE} : \drawUnitLine{EF}$ \byref{prop:VI.IV}. -Но $\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{DE} : \drawUnitLine{EF}$ \bycref{\hypref}. +Но $\drawUnitLine{AB} : \drawUnitLine{BC} :: \drawUnitLine{DE} : \drawUnitLine{EF}$ \byref{\hypref}. -$\therefore \drawUnitLine{DE} : \drawUnitLine{EF} :: \drawUnitLine{GE} : \drawUnitLine{EF}$, и~значит $\drawUnitLine{DE} = \drawUnitLine{GE}$ \bycref{prop:V.IX}. +$\therefore \drawUnitLine{DE} : \drawUnitLine{EF} :: \drawUnitLine{GE} : \drawUnitLine{EF}$, и~значит $\drawUnitLine{DE} = \drawUnitLine{GE}$ \byref{prop:V.IX}. Так же можно показать, что $\drawUnitLine{FD} = \drawUnitLine{FG}$. $\therefore$ у~двух треугольников с~общим основанием \drawUnitLine{EF} и~равными сторонами углы против равных сторон равны, т.~е. -$\drawAngle{DEF} = \drawAngle{GEF}$ и~$\drawAngle{EFD} = \drawAngle{EFG}$ \bycref{prop:I.VIII}. +$\drawAngle{DEF} = \drawAngle{GEF}$ и~$\drawAngle{EFD} = \drawAngle{EFG}$ \byref{prop:I.VIII}. -Но $\drawAngle{GEF} = \drawAngle{B}$ \bycref{\constref} и~$\therefore \drawAngle{DEF} = \drawAngle{B}$. +Но $\drawAngle{GEF} = \drawAngle{B}$ \byref{\constref} и~$\therefore \drawAngle{DEF} = \drawAngle{B}$. -По той же причине $\drawAngle{EFD} = \drawAngle{C}$, и~значит $\drawAngle{D} = \drawAngle{A}$ \bycref{prop:I.XXXII}, т.~е. треугольники равноугольны и~очевидно, что соответственные стороны располагаются между равными углами. +По той же причине $\drawAngle{EFD} = \drawAngle{C}$, и~значит $\drawAngle{D} = \drawAngle{A}$ \byref{prop:I.XXXII}, т.~е. треугольники равноугольны и~очевидно, что соответственные стороны располагаются между равными углами. \end{center} \qed @@ -11364,17 +11363,17 @@ \chapter*{Определения} \begin{center} Из концов \drawUnitLine{FD}, одной из сторон \triangleDEF\ при \drawAngle{EFD}, проведем \drawUnitLine{GD} и~\drawUnitLine{FG},\\ делая $\drawAngle{GFD} = \drawAngle{C}$ и~$\drawAngle{FDG} = \drawAngle{A}$,\\ -тогда $\drawAngle{G} = \drawAngle{B}$ \bycref{prop:I.XXXII}.\\ +тогда $\drawAngle{G} = \drawAngle{B}$ \byref{prop:I.XXXII}.\\ И поскольку два треугольника равноугольны,\\ -$\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{FG} : \drawUnitLine{FD}$ \bycref{prop:VI.IV},\\ -но $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \bycref{\hypref}, -$\therefore \drawUnitLine{FG} : \drawUnitLine{FD} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \bycref{prop:V.XI},\\ -и следовательно $\drawUnitLine{FG} = \drawUnitLine{EF}$ \bycref{prop:V.IX},\\ +$\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{FG} : \drawUnitLine{FD}$ \byref{prop:VI.IV},\\ +но $\drawUnitLine{BC} : \drawUnitLine{CA} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \byref{\hypref}, +$\therefore \drawUnitLine{FG} : \drawUnitLine{FD} :: \drawUnitLine{EF} : \drawUnitLine{FD}$ \byref{prop:V.XI},\\ +и следовательно $\drawUnitLine{FG} = \drawUnitLine{EF}$ \byref{prop:V.IX},\\ $\therefore \triangleDEF\ = -\drawLine[bottom][triangleDGF]{FD,FG,GD}$ во всех отношениях \bycref{prop:I.IV},\\ -но $\drawAngle{GFD} = \drawAngle{A}$ \bycref{\constref} и~$\therefore \drawAngle{EFD} = \drawAngle{A}$,\\ +\drawLine[bottom][triangleDGF]{FD,FG,GD}$ во всех отношениях \byref{prop:I.IV},\\ +но $\drawAngle{GFD} = \drawAngle{A}$ \byref{\constref} и~$\therefore \drawAngle{EFD} = \drawAngle{A}$,\\ и поскольку также $\drawAngle{EFD} = \drawAngle{C}$,\\ -$\drawAngle{E} = \drawAngle{B}$ \bycref{prop:I.XXXII}.\\ +$\drawAngle{E} = \drawAngle{B}$ \byref{prop:I.XXXII}.\\ И $\therefore$ \triangleABC\ и~\triangleDEF\ равноугольны с~равными углами против соответственных сторон. \end{center} @@ -11420,27 +11419,27 @@ \chapter*{Определения} \begin{center} Для начала допустим, что углы \drawAngle{A} и~\drawAngle{D} оба меньше прямого угла, тогда, если предположить, что \drawAngle{ABG,GBC} и~\drawAngle{E} между пропорциональными сторонами не равны, пусть \drawAngle{ABG,GBC} будет больше, и~сделаем $\drawAngle{GBC} = \drawAngle{E}$. -Поскольку $\drawAngle{C} = \drawAngle{F}$ \bycref{\hypref}\\ -и~$\drawAngle{GBC} = \drawAngle{E}$ \bycref{\constref},\\ -$\therefore \drawAngle{CGB} = \drawAngle{D}$ \bycref{prop:I.XXXII}. +Поскольку $\drawAngle{C} = \drawAngle{F}$ \byref{\hypref}\\ +и~$\drawAngle{GBC} = \drawAngle{E}$ \byref{\constref},\\ +$\therefore \drawAngle{CGB} = \drawAngle{D}$ \byref{prop:I.XXXII}. -$\therefore \drawUnitLine{BC} : \drawUnitLine{BG} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \bycref{prop:VI.IV},\\ -но $\drawUnitLine{BC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \bycref{\hypref}. +$\therefore \drawUnitLine{BC} : \drawUnitLine{BG} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \byref{prop:VI.IV},\\ +но $\drawUnitLine{BC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \byref{\hypref}. $\therefore \drawUnitLine{BC} : \drawUnitLine{BG} :: \drawUnitLine{BC} : \drawUnitLine{AB}$. -$\therefore \drawUnitLine{BG} = \drawUnitLine{AB}$ \bycref{prop:V.IX},\\ -и $\therefore \drawAngle{A} = \drawAngle{BGA}$ \bycref{prop:I.V}. +$\therefore \drawUnitLine{BG} = \drawUnitLine{AB}$ \byref{prop:V.IX},\\ +и $\therefore \drawAngle{A} = \drawAngle{BGA}$ \byref{prop:I.V}. -Но \drawAngle{A} меньше прямого угла \bycref{\hypref},\\ +Но \drawAngle{A} меньше прямого угла \byref{\hypref},\\ $\therefore$ \drawAngle{BGA} меньше прямого угла,\\ -и $\therefore$ \drawAngle{CGB} должен быть больше прямого угла \bycref{prop:I.XIII}, но он, как было показано, $= \drawAngle{D}$, и~значит меньше прямого угла, что невозможно. $\therefore$ \drawAngle{ABG,GBC} и~\drawAngle{E} не неравны. +и $\therefore$ \drawAngle{CGB} должен быть больше прямого угла \byref{prop:I.XIII}, но он, как было показано, $= \drawAngle{D}$, и~значит меньше прямого угла, что невозможно. $\therefore$ \drawAngle{ABG,GBC} и~\drawAngle{E} не неравны. -$\therefore$ они равны, и~поскольку $\drawAngle{C} = \drawAngle{F}$ \bycref{\hypref},\\ -$\therefore \drawAngle{A} = \drawAngle{D}$ \bycref{prop:I.XXXII}, и~следовательно треугольники равноугольны. +$\therefore$ они равны, и~поскольку $\drawAngle{C} = \drawAngle{F}$ \byref{\hypref},\\ +$\therefore \drawAngle{A} = \drawAngle{D}$ \byref{prop:I.XXXII}, и~следовательно треугольники равноугольны. \end{center} -А если предположить, что \drawAngle{A} и~\drawAngle{D} оба не меньше прямого угла, то можно доказать, как и~ранее, что треугольники равноугольны и~стороны при равных углах пропорциональны \bycref{prop:VI.IV}. +А если предположить, что \drawAngle{A} и~\drawAngle{D} оба не меньше прямого угла, то можно доказать, как и~ранее, что треугольники равноугольны и~стороны при равных углах пропорциональны \byref{prop:VI.IV}. \qed @@ -11467,11 +11466,11 @@ \chapter*{Определения} \problem[3]{Е}{сли}{из прямого угла в~прямоугольном треугольнике \drawPolygon[bottom][triangleABC]{ABD,ADC} проведен к~противоположной стороне перпендикуляр \drawUnitLine{AD}, треугольники \drawPolygon[bottom][triangleABD]{ABD} и~\drawPolygon[bottom][triangleADC]{ADC} по сторонам перпендикуляра подобны целому треугольнику и~друг другу.} \begin{center} -Поскольку $\drawAngle{DAB,CAD} = \drawAngle{D}$ \bycref{ax:I.XI}\\ +Поскольку $\drawAngle{DAB,CAD} = \drawAngle{D}$ \byref{ax:I.XI}\\ и~\drawAngle{B} общий \triangleABC\ и~\triangleABD,\\ -$\drawAngle{C} = \drawAngle{DAB}$ \bycref{prop:I.XXXII}. +$\drawAngle{C} = \drawAngle{DAB}$ \byref{prop:I.XXXII}. -$\therefore$ \triangleABC\ и~\triangleABD\ равноугольны, и~значит стороны при равных углах пропорциональны \bycref{prop:VI.IV}, и значит треугольники подобны \bycref{def:VI.I}. +$\therefore$ \triangleABC\ и~\triangleABD\ равноугольны, и~значит стороны при равных углах пропорциональны \byref{prop:VI.IV}, и значит треугольники подобны \byref{def:VI.I}. Так же можно доказать, что \triangleADC\ подобен \triangleABC, но \triangleABD, как было показано, тоже подобен \triangleABC, $\therefore$ \triangleABD\ и~\triangleADC\ подобны и~всему треугольнику, и~друг другу. \end{center} @@ -11517,12 +11516,12 @@ \chapter*{Определения} \drawSizedLine{AF} и~есть требуемая часть \drawSizedLine{AF,FB}. Поскольку $\drawSizedLine{DF} \parallel \drawSizedLine{BE}$,\\ -$\drawSizedLine{AF} : \drawSizedLine{FB} :: \drawSizedLine{AD} : \drawSizedLine{DC,CE}$ \bycref{prop:VI.II}. +$\drawSizedLine{AF} : \drawSizedLine{FB} :: \drawSizedLine{AD} : \drawSizedLine{DC,CE}$ \byref{prop:VI.II}. -И, присоединив \bycref{prop:V.XVIII},\\ +И, присоединив \byref{prop:V.XVIII},\\ $\drawSizedLine{AF,FB} : \drawSizedLine{AF} :: \drawSizedLine{AD,DC,CE} : \drawSizedLine{AD}$. -То есть \drawSizedLine{AD,DC,CE} содержит \drawSizedLine{AD} столько раз, сколько \drawSizedLine{AF,FB} содержит требуемую часть \bycref{\constref}. +То есть \drawSizedLine{AD,DC,CE} содержит \drawSizedLine{AD} столько раз, сколько \drawSizedLine{AF,FB} содержит требуемую часть \byref{\constref}. $\therefore$ \drawSizedLine{AF} и~есть требуемая часть. \end{center} @@ -11572,7 +11571,7 @@ \chapter*{Определения} \begin{center} Из любого конца \drawSizedLine{AF,FG,GB} проведем \drawSizedLine{AD,DE,EC} под любым углом. -Возьмем \drawSizedLine{AD}, \drawSizedLine{DE} и~\drawSizedLine{EC}, равные \drawSizedLine{A'D'}, \drawSizedLine{D'E'} и~\drawSizedLine{E'C'} соответственно \bycref{prop:I.II}. +Возьмем \drawSizedLine{AD}, \drawSizedLine{DE} и~\drawSizedLine{EC}, равные \drawSizedLine{A'D'}, \drawSizedLine{D'E'} и~\drawSizedLine{E'C'} соответственно \byref{prop:I.II}. Проведем \drawSizedLine{CB} и~проведем \drawSizedLine{EG} и~\drawSizedLine{DF} $\parallel$ ей. @@ -11582,10 +11581,10 @@ \chapter*{Определения} \nointerlineskip\hbox{\drawSizedLine{EG}} \nointerlineskip\hbox{\drawSizedLine{DF}}}\right\}$ все $\parallel$,\\ -$\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{EC} : \drawSizedLine{DE}$ \bycref{prop:VI.II},\\ -или $\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{E'C'} : \drawSizedLine{D'E'}$ \bycref{\constref},\\ -и $\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{DE} : \drawSizedLine{AD}$ \bycref{prop:VI.II},\\ -$\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{D'E'} : \drawSizedLine{A'D'}$ \bycref{\constref}. +$\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{EC} : \drawSizedLine{DE}$ \byref{prop:VI.II},\\ +или $\drawSizedLine{GB} : \drawSizedLine{FG} :: \drawSizedLine{E'C'} : \drawSizedLine{D'E'}$ \byref{\constref},\\ +и $\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{DE} : \drawSizedLine{AD}$ \byref{prop:VI.II},\\ +$\drawSizedLine{FG} : \drawSizedLine{AF} :: \drawSizedLine{D'E'} : \drawSizedLine{A'D'}$ \byref{\constref}. И $\therefore$ данная линия \drawSizedLine{AF,FG,GB} разделена подобно \drawSizedLine{A'D',D'E',E'C'}. \end{center} @@ -11629,15 +11628,15 @@ \chapter*{Определения} и~проведем \drawSizedLine{BC}. Сделаем $\drawSizedLine{BD} = \drawSizedLine{A'C'}$\\ -и проведем $\drawSizedLine{DE} \parallel \drawSizedLine{BC}$ \bycref{prop:I.XXXI}. +и проведем $\drawSizedLine{DE} \parallel \drawSizedLine{BC}$ \byref{prop:I.XXXI}. \drawSizedLine{CE} и~есть третья пропорциональная к~\drawSizedLine{AB} и~\drawSizedLine{A'C'}. Поскольку $\drawSizedLine{BC} \parallel \drawSizedLine{DE}$,\\ -$\therefore \drawSizedLine{AB} : \drawSizedLine{BD} :: \drawSizedLine{AC} : \drawSizedLine{CE}$ \bycref{prop:VI.II}. +$\therefore \drawSizedLine{AB} : \drawSizedLine{BD} :: \drawSizedLine{AC} : \drawSizedLine{CE}$ \byref{prop:VI.II}. -Но $\drawSizedLine{BD} = \drawSizedLine{AC} = \drawSizedLine{A'C'}$ \bycref{\constref};\\ -$\therefore \drawSizedLine{AB} : \drawSizedLine{A'C'} :: \drawSizedLine{A'C'} : \drawSizedLine{CE}$ \bycref{prop:V.VII}. +Но $\drawSizedLine{BD} = \drawSizedLine{AC} = \drawSizedLine{A'C'}$ \byref{\constref};\\ +$\therefore \drawSizedLine{AB} : \drawSizedLine{A'C'} :: \drawSizedLine{A'C'} : \drawSizedLine{CE}$ \byref{prop:V.VII}. \end{center} \qed @@ -11688,12 +11687,12 @@ \chapter*{Определения} возьмем $\drawUnitLine{GE} = \drawUnitLine{B}$,\\ возьмем $\drawUnitLine{DH} = \drawUnitLine{C}$,\\ проведем \drawUnitLine{GH}\\ -и $\drawUnitLine{EF} \parallel \drawUnitLine{GH}$ \bycref{prop:I.XXXI}. +и $\drawUnitLine{EF} \parallel \drawUnitLine{GH}$ \byref{prop:I.XXXI}. \drawSizedLine{HF} и~есть четвертая пропорциональная. Ведь, учитывая параллельные,\\ -$\drawUnitLine{DG} : \drawUnitLine{GE} :: \drawUnitLine{DH} : \drawUnitLine{HF}$ \bycref{prop:VI.II}. +$\drawUnitLine{DG} : \drawUnitLine{GE} :: \drawUnitLine{DH} : \drawUnitLine{HF}$ \byref{prop:VI.II}. Но $\left\{\vcenter{ @@ -11705,9 +11704,9 @@ \chapter*{Определения} \nointerlineskip\hbox{\drawSizedLine{DG}} \nointerlineskip\hbox{\drawSizedLine{GE}} \nointerlineskip\hbox{\drawSizedLine{DH}}}\right\}$ -\bycref{\constref}. +\byref{\constref}. -$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{C} : \drawUnitLine{HF}$ \bycref{prop:V.VII}. +$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{C} : \drawUnitLine{HF}$ \byref{prop:V.VII}. \end{center} \qed @@ -11769,10 +11768,10 @@ \chapter*{Определения} Проведем \drawUnitLine{AD} и~\drawUnitLine{CD}. -Поскольку \drawAngle{D} прямой угол \bycref{prop:III.XXXI}\\ +Поскольку \drawAngle{D} прямой угол \byref{prop:III.XXXI}\\ и \drawUnitLine{BD} это $\perp$ к~нему с~противной стороны,\\ -$\therefore$ \drawUnitLine{BD} и~есть средняя пропорциональная между \drawUnitLine{AB} и~\drawUnitLine{BC} \bycref{prop:VI.VIII},\\ -и $\therefore$ между \drawUnitLine{A} и~\drawUnitLine{B} \bycref{\constref}. +$\therefore$ \drawUnitLine{BD} и~есть средняя пропорциональная между \drawUnitLine{AB} и~\drawUnitLine{BC} \byref{prop:VI.VIII},\\ +и $\therefore$ между \drawUnitLine{A} и~\drawUnitLine{B} \byref{\constref}. \end{center} \qed @@ -11806,24 +11805,24 @@ \chapter*{Определения} \problem[3]{В}{ равных}{и равноугольных параллелограммах \drawPolygon{BFAD} и~\drawPolygon{CEBG} стороны при равных углах взаимно пропорциональны ($\drawUnitLine{BG} : \drawUnitLine{BF} :: \drawUnitLine{BD} : \drawUnitLine{BE}$). И~равноугольные параллелограммы, у~которых стороны взаимно пропорциональны, равны.} \begin{center} -Пусть \drawUnitLine{BG}, \drawUnitLine{BF}, \drawUnitLine{BD} и~\drawUnitLine{BE} будут так расположены, что \drawUnitLine{BG,BF} и~\drawUnitLine{BD,BE} составят прямые линии. Что они могут занять такое положение — очевидно \bycref{prop:I.XIII,prop:I.XIV,prop:I.XV}. +Пусть \drawUnitLine{BG}, \drawUnitLine{BF}, \drawUnitLine{BD} и~\drawUnitLine{BE} будут так расположены, что \drawUnitLine{BG,BF} и~\drawUnitLine{BD,BE} составят прямые линии. Что они могут занять такое положение — очевидно \byref{prop:I.XIII,prop:I.XIV,prop:I.XV}. Достроим \drawPolygon{EHFB}. Поскольку $\drawPolygon{CEBG} = \drawPolygon{BFAD}$,\\ -$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \bycref{prop:V.VII},\\ -$\therefore \drawUnitLine{BG} : \drawUnitLine{BF} :: \drawUnitLine{BD} : \drawUnitLine{BE}$ \bycref{prop:VI.I}. +$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \byref{prop:V.VII},\\ +$\therefore \drawUnitLine{BG} : \drawUnitLine{BF} :: \drawUnitLine{BD} : \drawUnitLine{BE}$ \byref{prop:VI.I}. При том же построении:\\ $\drawUnitLine{BG} : \drawUnitLine{BF} :: \left\{ \begin{aligned} - \drawPolygon{CEBG} &: \drawPolygon{EHFB} \mbox{\bycref{prop:VI.I}}\\ - \drawUnitLine{BD} &: \drawUnitLine{BE} \mbox{\bycref{\hypref}}\\ - \drawPolygon{BFAD} &: \drawPolygon{EHFB} \mbox{\bycref{prop:VI.I}}\\ + \drawPolygon{CEBG} &: \drawPolygon{EHFB} \mbox{\byref{prop:VI.I}}\\ + \drawUnitLine{BD} &: \drawUnitLine{BE} \mbox{\byref{\hypref}}\\ + \drawPolygon{BFAD} &: \drawPolygon{EHFB} \mbox{\byref{prop:VI.I}}\\ \end{aligned} \right.$ -$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \bycref{prop:V.XI},\\ -и $\therefore \drawPolygon{CEBG} = \drawPolygon{BFAD}$ \bycref{prop:V.IX}. +$\therefore \drawPolygon{CEBG} : \drawPolygon{EHFB} :: \drawPolygon{BFAD} : \drawPolygon{EHFB}$ \byref{prop:V.XI},\\ +и $\therefore \drawPolygon{CEBG} = \drawPolygon{BFAD}$ \byref{prop:V.IX}. \end{center} \qed @@ -11856,18 +11855,18 @@ \chapter*{Определения} И треугольники, имеющие по одному равному углу и~взаимно пропорциональные стороны при равных углах, равны.} \startsubproposition{Часть I.} -Пусть треугольники будут расположены так, что равные углы \drawAngle{DAE} и~\drawAngle{BAC} будут вертикальны, то есть так, что \drawUnitLine{AE} и~\drawUnitLine{BA} будут на одной прямой. Таким образом и~\drawUnitLine{AC} с~\drawUnitLine{DA} будут на одной прямой \bycref{prop:I.XIV}. +Пусть треугольники будут расположены так, что равные углы \drawAngle{DAE} и~\drawAngle{BAC} будут вертикальны, то есть так, что \drawUnitLine{AE} и~\drawUnitLine{BA} будут на одной прямой. Таким образом и~\drawUnitLine{AC} с~\drawUnitLine{DA} будут на одной прямой \byref{prop:I.XIV}. \begin{center} Проведем \drawUnitLine{BD}, тогда\\ $\begin{aligned} \drawUnitLine{DA} : \drawUnitLine{BA} - &:: \drawPolygon{DAE} : \drawPolygon{DAB} \mbox{\bycref{prop:VI.I}}\\ - &:: \drawPolygon{ABC} : \drawPolygon{DAB} \mbox{\bycref{prop:V.VII}}\\ - &:: \drawUnitLine{AC} : \drawUnitLine{DA} \mbox{\bycref{prop:VI.I}}\\ + &:: \drawPolygon{DAE} : \drawPolygon{DAB} \mbox{\byref{prop:VI.I}}\\ + &:: \drawPolygon{ABC} : \drawPolygon{DAB} \mbox{\byref{prop:V.VII}}\\ + &:: \drawUnitLine{AC} : \drawUnitLine{DA} \mbox{\byref{prop:VI.I}}\\ \end{aligned}$ -$\therefore \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$ \bycref{prop:V.XI}. +$\therefore \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$ \byref{prop:V.XI}. \end{center} \vfill\pagebreak @@ -11876,14 +11875,14 @@ \chapter*{Определения} \begin{center} При том же построении. -$\drawPolygon{DAE} : \drawPolygon{DAB} :: \drawUnitLine{DA} : \drawUnitLine{BA}$ \bycref{prop:VI.I}\\ -и $\drawUnitLine{AC} : \drawUnitLine{DA} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \bycref{prop:VI.I}. +$\drawPolygon{DAE} : \drawPolygon{DAB} :: \drawUnitLine{DA} : \drawUnitLine{BA}$ \byref{prop:VI.I}\\ +и $\drawUnitLine{AC} : \drawUnitLine{DA} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \byref{prop:VI.I}. -Но $ \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$, \bycref{\hypref}. +Но $ \drawUnitLine{AE} : \drawUnitLine{BA} :: \drawUnitLine{AC} : \drawUnitLine{DA}$, \byref{\hypref}. -$\therefore \drawPolygon{DAE} : \drawPolygon{DAB} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \bycref{prop:V.XI}. +$\therefore \drawPolygon{DAE} : \drawPolygon{DAB} :: \drawPolygon{ABC} : \drawPolygon{DAB}$ \byref{prop:V.XI}. -$\therefore \drawPolygon{DAE} = \drawPolygon{ABC}$ \bycref{prop:V.IX}. +$\therefore \drawPolygon{DAE} = \drawPolygon{ABC}$ \byref{prop:V.IX}. \end{center} \qed @@ -11934,10 +11933,10 @@ \chapter*{Определения} Достроим параллелограммы \drawPolygon{ABFG} и~\drawPolygon{CDEH}. И поскольку\\ -$\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \bycref{\hypref},\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \bycref{\constref}. +$\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \byref{\hypref},\\ +$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \byref{\constref}. -$\therefore \drawPolygon{ABFG} = \drawPolygon{CDEH}$ \bycref{prop:VI.XIV},\\ +$\therefore \drawPolygon{ABFG} = \drawPolygon{CDEH}$ \byref{prop:VI.XIV},\\ то есть прямоугольник, заключенный между крайними, равен прямоугольнику, заключенному между средними. \end{center} @@ -11949,12 +11948,12 @@ \chapter*{Определения} Поскольку $\drawUnitLine{F} = \drawUnitLine{AG}$, $\drawPolygon{ABFG} = \drawPolygon{CDEH}$\\ и $\drawUnitLine{CH} = \drawUnitLine{E}$,\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \bycref{prop:VI.XIV}. +$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{CH} : \drawUnitLine{AG}$ \byref{prop:VI.XIV}. Но $\drawUnitLine{CH} = \drawUnitLine{E}$\\ -и $\drawUnitLine{AG} = \drawUnitLine{F}$ \bycref{\constref}. +и $\drawUnitLine{AG} = \drawUnitLine{F}$ \byref{\constref}. -$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \bycref{prop:V.VII}. +$\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{E} : \drawUnitLine{F}$ \byref{prop:V.VII}. \end{center} \qed @@ -12006,7 +12005,7 @@ \chapter*{Определения} \mbox{ то } \drawUnitLine{A} : \drawUnitLine{B} &:: \drawUnitLine{D} : \drawUnitLine{C}, \\ \therefore \drawUnitLine{A} \times \drawUnitLine{C} &= \drawUnitLine{B} \times \drawUnitLine{D} \\ \end{aligned}$\\ -\bycref{prop:VI.XVI}. +\byref{prop:VI.XVI}. Но $\drawUnitLine{D} = \drawUnitLine{B}$,\\ $\therefore \drawUnitLine{B} \times \drawUnitLine{D} = \drawUnitLine{B} \times \drawUnitLine{B} \mbox{ или } = \drawUnitLine{B}^2$,\\ @@ -12018,7 +12017,7 @@ \chapter*{Определения} Допустим, $\drawUnitLine{D} = \drawUnitLine{B}$,\\ тогда $\drawUnitLine{A} \times \drawUnitLine{C} = \drawUnitLine{D} \times \drawUnitLine{B}$. -$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{D} : \drawUnitLine{C}$ \bycref{prop:VI.XVI}\\ +$\therefore \drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{D} : \drawUnitLine{C}$ \byref{prop:VI.XVI}\\ и $\drawUnitLine{A} : \drawUnitLine{B} :: \drawUnitLine{B} : \drawUnitLine{C}$. \end{center} @@ -12090,19 +12089,19 @@ \chapter*{Определения} Тогда $\polygonCD = \drawPolygon[middle][polygonAB]{BAG,BGH,BHK}$. -Из построения и~\bycref{prop:I.XXXII} очевидно, что фигуры равноугольны, а~поскольку треугольники \drawPolygon{DCF} и~\drawPolygon{BAG} равноугольны,\\ -то, согласно \bycref{prop:VI.IV}, +Из построения и~\byref{prop:I.XXXII} очевидно, что фигуры равноугольны, а~поскольку треугольники \drawPolygon{DCF} и~\drawPolygon{BAG} равноугольны,\\ +то, согласно \byref{prop:VI.IV}, $\drawUnitLine{AB}:\drawUnitLine{AG} :: \drawUnitLine{CD} : \drawUnitLine{CF}$\\ и $\drawUnitLine{AG}:\drawUnitLine{BG} :: \drawUnitLine{CF} : \drawUnitLine{DF}$. И поскольку \drawPolygon{DFE} и~\drawPolygon{BGH} равноугольны,\\ $\drawUnitLine{BG}:\drawUnitLine{GH} :: \drawUnitLine{DF} : \drawUnitLine{FE}$\\ $\therefore$ по равенству,\\ -$\drawUnitLine{AG}:\drawUnitLine{GH} :: \drawUnitLine{CF} : \drawUnitLine{FE}$ \bycref{prop:V.XXII}. +$\drawUnitLine{AG}:\drawUnitLine{GH} :: \drawUnitLine{CF} : \drawUnitLine{FE}$ \byref{prop:V.XXII}. Тем же способом можно показать, что оставшиеся стороны фигуры пропорциональны. -$\therefore$ \bycref{prop:VI.I}\\ +$\therefore$ \byref{prop:VI.I}\\ \polygonAB\ подобна \polygonCD\ и~расположена на данной прямой \drawUnitLine{AB}. \end{center} @@ -12148,22 +12147,22 @@ \chapter*{Определения} $\drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{EF} : \drawUnitLine{BG}$. Проведем \drawUnitLine{AG}.\\ -$\drawUnitLine{BG,GC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \bycref{prop:VI.IV}. +$\drawUnitLine{BG,GC} : \drawUnitLine{AB} :: \drawUnitLine{EF} : \drawUnitLine{DE}$ \byref{prop:VI.IV}. -$\therefore \drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{AB} : \drawUnitLine{DE}$ \bycref{prop:V.XVI}. +$\therefore \drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{AB} : \drawUnitLine{DE}$ \byref{prop:V.XVI}. -Но $\drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{EF} : \drawUnitLine{BG}$ \bycref{\constref},\\ +Но $\drawUnitLine{BG,GC} : \drawUnitLine{EF} :: \drawUnitLine{EF} : \drawUnitLine{BG}$ \byref{\constref},\\ $\therefore \drawUnitLine{EF} : \drawUnitLine{BG} :: \drawUnitLine{AB} : \drawUnitLine{DE}$. -Следовательно, $\triangleDEF = \drawPolygon[bottom][triangleABG]{ABG}$, поскольку у~них стороны при равных углах \drawAngle{E} и~\drawAngle{B} взаимно пропорциональны \bycref{prop:VI.XV}. +Следовательно, $\triangleDEF = \drawPolygon[bottom][triangleABG]{ABG}$, поскольку у~них стороны при равных углах \drawAngle{E} и~\drawAngle{B} взаимно пропорциональны \byref{prop:VI.XV}. -$\therefore \triangleABC : \triangleDEF :: \triangleABC : \triangleABG$ \bycref{prop:V.VII}. +$\therefore \triangleABC : \triangleDEF :: \triangleABC : \triangleABG$ \byref{prop:V.VII}. -Но $\triangleABC : \triangleABG :: \drawUnitLine{BG,GC} : \drawUnitLine{BG}$ \bycref{prop:VI.I}. +Но $\triangleABC : \triangleABG :: \drawUnitLine{BG,GC} : \drawUnitLine{BG}$ \byref{prop:VI.I}. $\therefore \triangleABC : \triangleDEF :: \drawUnitLine{BG,GC} : \drawUnitLine{BG}$. -То есть эти треугольники относятся друг к~другу в~двойном отношении соответственных сторон \drawUnitLine{EF} и~\drawUnitLine{BG,GC} \bycref{def:V.XI}. +То есть эти треугольники относятся друг к~другу в~двойном отношении соответственных сторон \drawUnitLine{EF} и~\drawUnitLine{BG,GC} \byref{def:V.XI}. \end{center} \qed @@ -12220,7 +12219,7 @@ \chapter*{Определения} Проведя \drawUnitLine{BE}, \drawUnitLine{BD}, \drawUnitLine{GL} и~\drawUnitLine{GK}, разделим многоугольники на треугольники. Поскольку многоугольники подобны $\drawAngle{C} = \drawAngle{H}$, и~$\drawUnitLine{BC} : \drawUnitLine{CD} :: \drawUnitLine{GH} : \drawUnitLine{HK}$. \begin{center} -$\therefore$ \drawPolygon{BCD} и~\drawPolygon{GHK} подобны и~$\drawAngle{BDC} = \drawAngle{GKH}$ \bycref{prop:VI.VI},\\ +$\therefore$ \drawPolygon{BCD} и~\drawPolygon{GHK} подобны и~$\drawAngle{BDC} = \drawAngle{GKH}$ \byref{prop:VI.VI},\\ но $\drawAngle{BDE,BDC} = \drawAngle{GKL,GKH}$, поскольку они являются углами подобных многоугольников. Следовательно, остатки \drawAngle{BDE} и~\drawAngle{GKL} равны. А значит $\drawUnitLine{BD} : \drawUnitLine{CD} :: \drawUnitLine{GK} : \drawUnitLine{HK}$,\\ @@ -12229,14 +12228,14 @@ \chapter*{Определения} учитывая подобие многоугольников. $\therefore \drawUnitLine{BD} : \drawUnitLine{DE} :: \drawUnitLine{GK} : \drawUnitLine{KL}$\\ -по равенству \bycref{prop:V.XXII}, и~поскольку пропорциональные стороны заключают равные углы, треугольники \drawPolygon{BDE} и~\drawPolygon{GKL} подобны \bycref{prop:VI.VI}. +по равенству \byref{prop:V.XXII}, и~поскольку пропорциональные стороны заключают равные углы, треугольники \drawPolygon{BDE} и~\drawPolygon{GKL} подобны \byref{prop:VI.VI}. Так же можно показать, что и~треугольники \drawPolygon{BEA} и~\drawPolygon{GLF} подобны. -Но \drawPolygon{BCD} имеет к~\drawPolygon{GHK} двойное отношение \drawUnitLine{BD} к~\drawUnitLine{GK} \bycref{prop:VI.XIX},\\ +Но \drawPolygon{BCD} имеет к~\drawPolygon{GHK} двойное отношение \drawUnitLine{BD} к~\drawUnitLine{GK} \byref{prop:VI.XIX},\\ и \drawPolygon{BDE} к~\drawPolygon{GKL} точно так же имеет двойное отношение \drawUnitLine{BD} к~\drawUnitLine{GK}. -$\therefore \drawPolygon{BCD} : \drawPolygon{GHK} :: \drawPolygon{BDE} : \drawPolygon{GKL}$ \bycref{prop:V.XI}. +$\therefore \drawPolygon{BCD} : \drawPolygon{GHK} :: \drawPolygon{BDE} : \drawPolygon{GKL}$ \byref{prop:V.XI}. Теперь \drawPolygon{BDE} к~\drawPolygon{GKL} имеет двойное отношение \drawUnitLine{BE} к~\drawUnitLine{GL}, и~\drawPolygon{BEA} к~\drawPolygon{GLF} двойное отношение \drawUnitLine{BE} к~\drawUnitLine{GL}. @@ -12246,7 +12245,7 @@ \chapter*{Определения} \end{aligned}$. \end{center} -И поскольку как относится одно из предыдущих к~одному из последующих, так и~все предыдущие вместе ко всем последующим вместе, подобные треугольники относятся друг к~другу, как многоугольники целиком \bycref{prop:V.XII}. +И поскольку как относится одно из предыдущих к~одному из последующих, так и~все предыдущие вместе ко всем последующим вместе, подобные треугольники относятся друг к~другу, как многоугольники целиком \byref{prop:V.XII}. \begin{center} Но \drawPolygon{BCD} к~\drawPolygon{GHK} имеет двойное отношение \drawUnitLine{BC} к~\drawUnitLine{GH}. @@ -12281,7 +12280,7 @@ \chapter*{Определения} \drawCurrentPictureInMargin \problem{Ф}{игуры}{\drawPolygon[bottom]{A} и~\drawPolygon[bottom]{B}, подобные одной фигуре \drawPolygon[bottom]{C}, подобны друг другу.} -Поскольку \polygonA\ и~\polygonC\ подобны, они равноугольны, и~стороны при равных углах у~них пропорциональны \bycref{def:VI.I}. И~поскольку \polygonB\ и~\polygonC\ также подобны, и~стороны при равных углах у~них пропорциональны, \polygonA\ и~\polygonB\ тоже равноугольны и~у них тоже стороны при равных углах пропорциональны \bycref{prop:V.XI}, и~значит они тоже подобны. +Поскольку \polygonA\ и~\polygonC\ подобны, они равноугольны, и~стороны при равных углах у~них пропорциональны \byref{def:VI.I}. И~поскольку \polygonB\ и~\polygonC\ также подобны, и~стороны при равных углах у~них пропорциональны, \polygonA\ и~\polygonB\ тоже равноугольны и~у них тоже стороны при равных углах пропорциональны \byref{prop:V.XI}, и~значит они тоже подобны. \qed @@ -12333,17 +12332,17 @@ \chapter*{Определения} \startsubproposition{Часть I.} \begin{center} -Возьмем \drawUnitLine{O}, третью пропорциональную к~\drawUnitLine{AB} и~\drawUnitLine{CD}, и~\drawUnitLine{P}, третью пропорциональную к~\drawUnitLine{EF} и~\drawUnitLine{GH} \bycref{prop:VI.XI}. +Возьмем \drawUnitLine{O}, третью пропорциональную к~\drawUnitLine{AB} и~\drawUnitLine{CD}, и~\drawUnitLine{P}, третью пропорциональную к~\drawUnitLine{EF} и~\drawUnitLine{GH} \byref{prop:VI.XI}. -Поскольку $\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \bycref{\hypref},\\ -$\drawUnitLine{CD} : \drawUnitLine{O} :: \drawUnitLine{GH} : \drawUnitLine{P}$ \bycref{\constref}. +Поскольку $\drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \byref{\hypref},\\ +$\drawUnitLine{CD} : \drawUnitLine{O} :: \drawUnitLine{GH} : \drawUnitLine{P}$ \byref{\constref}. $\therefore$ по равенству\\ $\drawUnitLine{AB} : \drawUnitLine{O} :: \drawUnitLine{EF} : \drawUnitLine{P}$,\\ -но $\drawPolygon{AB} : \drawPolygon{CD} :: \drawUnitLine{AB} : \drawUnitLine{O}$ \bycref{prop:VI.XX}\\ +но $\drawPolygon{AB} : \drawPolygon{CD} :: \drawUnitLine{AB} : \drawUnitLine{O}$ \byref{prop:VI.XX}\\ и $\drawPolygon{EF} : \drawPolygon{GH} :: \drawUnitLine{EF} : \drawUnitLine{P}$. -$\therefore \drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \bycref{prop:V.XI}. +$\therefore \drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \byref{prop:V.XI}. \end{center} \vfill\pagebreak @@ -12351,11 +12350,11 @@ \chapter*{Определения} \startsubproposition{Часть II.} \begin{center} Оставим то же построение.\\ -$\drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \bycref{\hypref}. +$\drawPolygon{AB} : \drawPolygon{CD} :: \drawPolygon{EF} : \drawPolygon{GH}$ \byref{\hypref}. -$\therefore \drawUnitLine{AB} : \drawUnitLine{O} :: \drawUnitLine{EF} : \drawUnitLine{P}$ \bycref{\constref}. +$\therefore \drawUnitLine{AB} : \drawUnitLine{O} :: \drawUnitLine{EF} : \drawUnitLine{P}$ \byref{\constref}. -И $\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \bycref{prop:V.XI}. +И $\therefore \drawUnitLine{AB} : \drawUnitLine{CD} :: \drawUnitLine{EF} : \drawUnitLine{GH}$ \byref{prop:V.XI}. \end{center} \qed @@ -12395,14 +12394,14 @@ \chapter*{Определения} \begin{center} Поскольку $\drawAngle{BCD} + \drawAngle{DCG} = \drawTwoRightAngles$\\ -и $\drawAngle{GCE} = \drawAngle{BCD}$ \bycref{\hypref},\\ +и $\drawAngle{GCE} = \drawAngle{BCD}$ \byref{\hypref},\\ $\drawAngle{GCE} + \drawAngle{DCG} = \drawTwoRightAngles$,\\ -и $\therefore$ \drawUnitLine{CE} и~\drawUnitLine{DC} образуют одну прямую \bycref{prop:I.XIV}. +и $\therefore$ \drawUnitLine{CE} и~\drawUnitLine{DC} образуют одну прямую \byref{prop:I.XIV}. Достроим \drawPolygon[middle]{CDHG}. -Поскольку $\polygonABCD\ : \polygonCDHG\ :: \drawUnitLine{BC} : \drawUnitLine{CG}$ \bycref{prop:VI.I}\\ -и $\polygonCDHG\ : \polygonEFGC\ :: \drawUnitLine{DC} : \drawUnitLine{CE}$ \bycref{prop:VI.I},\\ +Поскольку $\polygonABCD\ : \polygonCDHG\ :: \drawUnitLine{BC} : \drawUnitLine{CG}$ \byref{prop:VI.I}\\ +и $\polygonCDHG\ : \polygonEFGC\ :: \drawUnitLine{DC} : \drawUnitLine{CE}$ \byref{prop:VI.I},\\ \polygonABCD\ имеет к~\polygonEFGC\ отношение, составленное из отношений \drawUnitLine{BC} к~\drawUnitLine{CG} и~\drawUnitLine{DC} к~\drawUnitLine{CE}. \end{center} @@ -12481,7 +12480,7 @@ \chapter*{Определения} stopAutoLabeling; stopGlobalRotation; } -подобны \bycref{prop:VI.IV},\\ +подобны \byref{prop:VI.IV},\\ $\therefore \drawUnitLine{GA} : \drawUnitLine{GF} :: \drawUnitLine{GA,DG} : \drawUnitLine{DK,KC}$, а оставшиеся противоположные стороны равны им. @@ -12561,25 +12560,25 @@ \chapter*{Определения} stopAutoLabeling; stopTempAngleScale; } = \polygonD$\\ -с $\drawAngle{B} = \drawAngle{C}$ \bycref{prop:I.XLV},\\ -и тогда \drawUnitLine{BC} и~\drawUnitLine{CF} будут лежать на одной прямой \bycref{prop:I.XXIX,prop:I.XIV}. +с $\drawAngle{B} = \drawAngle{C}$ \byref{prop:I.XLV},\\ +и тогда \drawUnitLine{BC} и~\drawUnitLine{CF} будут лежать на одной прямой \byref{prop:I.XXIX,prop:I.XIV}. -Найдем между \drawUnitLine{BC} и~\drawUnitLine{CF} среднюю пропорциональную \drawUnitLine{GH} \bycref{prop:VI.XIII},\\ +Найдем между \drawUnitLine{BC} и~\drawUnitLine{CF} среднюю пропорциональную \drawUnitLine{GH} \byref{prop:VI.XIII},\\ и на \drawUnitLine{GH} опишем \drawPolygon[middle][triangleKGH]{KGH},\\ подобный \triangleABC\ и~подобно расположенный. Тогда $\triangleKGH\ = \polygonD$. Поскольку \triangleABC\ и~\triangleKGH\ подобны\\ -и $\drawUnitLine{BC} : \drawUnitLine{GH} :: \drawUnitLine{GH} : \drawUnitLine{CF}$ \bycref{\constref},\\ -$\triangleABC\ : \triangleKGH\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \bycref{prop:VI.XX}. +и $\drawUnitLine{BC} : \drawUnitLine{GH} :: \drawUnitLine{GH} : \drawUnitLine{CF}$ \byref{\constref},\\ +$\triangleABC\ : \triangleKGH\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \byref{prop:VI.XX}. -Но $\polygonBCEL\ : \polygonCFME\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \bycref{prop:VI.I}. +Но $\polygonBCEL\ : \polygonCFME\ :: \drawUnitLine{BC} : \drawUnitLine{CF}$ \byref{prop:VI.I}. -$\therefore \triangleABC\ : \triangleKGH\ :: \polygonBCEL\ : \polygonCFME$ \bycref{prop:V.XI}. +$\therefore \triangleABC\ : \triangleKGH\ :: \polygonBCEL\ : \polygonCFME$ \byref{prop:V.XI}. -Но $\triangleABC\ = \polygonBCEL$ \bycref{\constref},\\ -и $\therefore \triangleKGH\ = \polygonCFME$ \bycref{prop:V.XIV};\\ -и $\polygonCFME\ = \polygonD$ \bycref{\constref}. +Но $\triangleABC\ = \polygonBCEL$ \byref{\constref},\\ +и $\therefore \triangleKGH\ = \polygonCFME$ \byref{prop:V.XIV};\\ +и $\polygonCFME\ = \polygonD$ \byref{\constref}. И, следовательно, \triangleKGH, подобный \triangleABC, вместе с~тем $= \polygonD$. \end{center} @@ -12650,15 +12649,15 @@ \chapter*{Определения} draw byLabelsOnPolygon(A, H, C, noPoint)(ALL_LABELS, 0); stopGlobalRotation; } -будет диагональю \parallelogramABCD, и~проведем $\drawUnitLine{KH} \parallel \drawUnitLine{AG}$ \bycref{prop:I.XXXI}. +будет диагональю \parallelogramABCD, и~проведем $\drawUnitLine{KH} \parallel \drawUnitLine{AG}$ \byref{prop:I.XXXI}. -Поскольку \drawLine[middle][parallelogramAKHG]{AG,GH,KH,AK} и~\parallelogramABCD\ на одной диагонали \lineAHC\ и~\drawAngle{A} уних общий, они подобны \bycref{prop:VI.XXIV}. +Поскольку \drawLine[middle][parallelogramAKHG]{AG,GH,KH,AK} и~\parallelogramABCD\ на одной диагонали \lineAHC\ и~\drawAngle{A} уних общий, они подобны \byref{prop:VI.XXIV}. $\therefore \drawSizedLine{AG} : \drawSizedLine{AK} :: \drawSizedLine{AG,GD} : \drawSizedLine{AK,KE,EB}$;\\ -но $\drawSizedLine{AG} : \drawSizedLine{AK,KE} :: \drawSizedLine{AG,GD} : \drawSizedLine{AK,KE,EB}$ \bycref{\hypref}. +но $\drawSizedLine{AG} : \drawSizedLine{AK,KE} :: \drawSizedLine{AG,GD} : \drawSizedLine{AK,KE,EB}$ \byref{\hypref}. $\therefore \drawSizedLine{AG} : \drawSizedLine{AK} :: \drawSizedLine{AG} : \drawSizedLine{AK,KE}$,\\ -и $\therefore \drawSizedLine{AK} = \drawSizedLine{AK,KE}$ \bycref{prop:V.IX}, что невозможно. +и $\therefore \drawSizedLine{AK} = \drawSizedLine{AK,KE}$ \byref{prop:V.IX}, что невозможно. $\therefore$ \lineAHC\ не является диагональю \parallelogramABCD, и~таким же образом можно показать, что диагональю не является никакая другая прямая, кроме \drawUnitLine{AC}. \end{center} @@ -12700,7 +12699,7 @@ \chapter*{Определения} Тогда $\drawPolygon[middle]{ADGKEF,CDGK} > \drawPolygon[middle]{BCKH,CDGK}$. \end{center} -Поскольку, как уже было показано \bycref{prop:II.V}, квадрат половины линии равен прямоугольнику, заключенному между неравными частями вместе с~квадратом на части между серединой и~точкой неравного рассечения. Квадрат, описанный на половине линии, превосходит, таким образом, прямоугольник, заключенный между неравными частями линии. +Поскольку, как уже было показано \byref{prop:II.V}, квадрат половины линии равен прямоугольнику, заключенному между неравными частями вместе с~квадратом на части между серединой и~точкой неравного рассечения. Квадрат, описанный на половине линии, превосходит, таким образом, прямоугольник, заключенный между неравными частями линии. \qed @@ -12745,21 +12744,21 @@ \chapter*{Определения} и если $\drawSizedLine{AE,EC}^2 = \drawSizedLine{G}^2$, то задача решена. Но если $\drawSizedLine{AE,EC}^2 \neq \drawSizedLine{G}^2$,\\ -тогда $\drawSizedLine{AE,EC} > \drawSizedLine{G}$ \bycref{\hypref}. +тогда $\drawSizedLine{AE,EC} > \drawSizedLine{G}$ \byref{\hypref}. Проведем $\drawSizedLine{DC} \perp \drawSizedLine{AE,EC} = \drawSizedLine{G}$,\\ сделаем $\drawSizedLine{DC,CF} = \drawSizedLine{AE,EC} \mbox{ или } \drawSizedLine{CB}$,\\ с \drawSizedLine{DC,CF} в~качестве радиуса опишем круг, секущий данную прямую, проведем \drawSizedLine{DE}. -Тогда $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} + \drawSizedLine{EC}^2 = \drawSizedLine{AE,EC}^2$ \bycref{prop:II.V} $= \drawSizedLine{DE}^2$. +Тогда $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} + \drawSizedLine{EC}^2 = \drawSizedLine{AE,EC}^2$ \byref{prop:II.V} $= \drawSizedLine{DE}^2$. -Но $\drawSizedLine{DE}^2 = \drawSizedLine{DC}^2 + \drawSizedLine{EC}^2$ \bycref{prop:I.XLVII}; +Но $\drawSizedLine{DE}^2 = \drawSizedLine{DC}^2 + \drawSizedLine{EC}^2$ \byref{prop:I.XLVII}; $\therefore \drawSizedLine{AE} \times \drawSizedLine{EC,CB} + \drawSizedLine{EC}^2 = \drawSizedLine{DC}^2 + \drawSizedLine{EC}^2$,\\ из обеих вычтем $\drawSizedLine{EC}^2$,\\ и $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} = \drawSizedLine{DC}^2$. -Но $\drawSizedLine{DC} = \drawSizedLine{G}$ \bycref{\constref},\\ +Но $\drawSizedLine{DC} = \drawSizedLine{G}$ \byref{\constref},\\ и $\therefore$ \drawSizedLine{AE,EC,CB} рассечена так, что $\drawSizedLine{AE} \times \drawSizedLine{EC,CB} = \drawSizedLine{G}^2$. \end{center} @@ -12799,9 +12798,9 @@ \chapter*{Определения} проведем \drawSizedLine{CD},\\ и с~\drawSizedLine{CD} в~качестве радиуса опишем круг, пересекающий продленную \drawSizedLine{AC,CB}. -Тогда $\drawSizedLine{AC,CB,BF} \times \drawSizedLine{BF} + \drawSizedLine{CB}^2 = \drawSizedLine{AC,CB}^2$ \bycref{prop:II.VI} $= \drawSizedLine{CD}^2$. +Тогда $\drawSizedLine{AC,CB,BF} \times \drawSizedLine{BF} + \drawSizedLine{CB}^2 = \drawSizedLine{AC,CB}^2$ \byref{prop:II.VI} $= \drawSizedLine{CD}^2$. -Но $\drawSizedLine{CD}^2 = \drawSizedLine{BD}^2 + \drawSizedLine{CB}^2$ \bycref{prop:I.XLVII}. +Но $\drawSizedLine{CD}^2 = \drawSizedLine{BD}^2 + \drawSizedLine{CB}^2$ \byref{prop:I.XLVII}. $\therefore \drawSizedLine{AC,CB,BF} \times \drawSizedLine{BF} + \drawSizedLine{CB}^2 = \drawSizedLine{BD}^2 + \drawSizedLine{CB}^2$,\\ из обеих вычтем $\drawSizedLine{CB}^2$,\\ @@ -12848,13 +12847,13 @@ \chapter*{Определения} \problem{Р}{ассечь}{данную прямую \drawProportionalLine{AE,EB} в~крайнем и~среднем отношении.} \begin{center} -На \drawProportionalLine{AE,EB} опишем квадрат \drawPolygon[middle][squareABHC]{ACFE,EFHB} \bycref{prop:I.XLVI}. +На \drawProportionalLine{AE,EB} опишем квадрат \drawPolygon[middle][squareABHC]{ACFE,EFHB} \byref{prop:I.XLVI}. -И продлим \drawProportionalLine{CA} так, что $\drawProportionalLine{CA,AG} \times \drawProportionalLine{AG} = \drawProportionalLine{AE,EB}^2$ \bycref{prop:VI.XXIX}. +И продлим \drawProportionalLine{CA} так, что $\drawProportionalLine{CA,AG} \times \drawProportionalLine{AG} = \drawProportionalLine{AE,EB}^2$ \byref{prop:VI.XXIX}. Возьмем $\drawProportionalLine{AE} = \drawProportionalLine{AG}$\\ и проведем $\drawProportionalLine{DE} \parallel \drawProportionalLine{CA,AG}$,\\ -встречающуюся с~$\drawProportionalLine{GD} \parallel \drawProportionalLine{AE,EB}$ \bycref{prop:I.XXXI}. +встречающуюся с~$\drawProportionalLine{GD} \parallel \drawProportionalLine{AE,EB}$ \byref{prop:I.XXXI}. Тогда $\drawFromCurrentPicture[middle][rectangleCFDG]{ draw byNamedPolygon(ACFE); @@ -12870,7 +12869,7 @@ \chapter*{Определения} То есть $\drawProportionalLine{AE}^2 = \drawProportionalLine{AE,EB} \times \drawProportionalLine{EB}$. $\therefore \drawProportionalLine{AE,EB} : \drawProportionalLine{AE} :: \drawProportionalLine{AE} : \drawProportionalLine{EB}$,\\ -и \drawProportionalLine{AE,EB} разделена в~крайнем и~среднем отношении \bycref{def:VI.III}. +и \drawProportionalLine{AE,EB} разделена в~крайнем и~среднем отношении \byref{def:VI.III}. \end{center} \qed @@ -12913,7 +12912,7 @@ \chapter*{Определения} \begin{center} Из прямого угла проведем перпендикуляр \drawProportionalLine{AD} к~\drawProportionalLine{BD,DC}. -Тогда $\drawProportionalLine{BD,DC} : \drawProportionalLine{CA} :: \drawProportionalLine{CA} : \drawProportionalLine{DC}$ \bycref{prop:VI.VIII}. +Тогда $\drawProportionalLine{BD,DC} : \drawProportionalLine{CA} :: \drawProportionalLine{CA} : \drawProportionalLine{DC}$ \byref{prop:VI.VIII}. $\therefore \drawFromCurrentPicture[bottom][figBC]{ @@ -12929,7 +12928,7 @@ \chapter*{Определения} draw byNamedPolygon(CA); stopAutoLabeling; stopGlobalRotation; -} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{DC}$ \bycref{prop:VI.XX}. +} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{DC}$ \byref{prop:VI.XX}. Но $\figBC\ : \drawFromCurrentPicture[bottom][figAB]{ @@ -12938,7 +12937,7 @@ \chapter*{Определения} draw byNamedPolygon(AB); stopAutoLabeling; stopGlobalRotation; -} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{BD}$ \bycref{prop:VI.XX}. +} :: \drawProportionalLine{BD,DC} : \drawProportionalLine{BD}$ \byref{prop:VI.XX}. Значит, $\drawProportionalLine{DC} + \drawProportionalLine{BD} : \drawProportionalLine{BD,DC} :: \figAB\ + \figCA\ : \figBC$. @@ -13003,21 +13002,21 @@ \chapter*{Определения} \begin{center} Поскольку $\drawUnitLine{AB} \parallel \drawUnitLine{DC}$,\\ -$\drawAngle{A} = \drawAngle{ACD}$ \bycref{prop:I.XXIX}. +$\drawAngle{A} = \drawAngle{ACD}$ \byref{prop:I.XXIX}. И поскольку $\drawUnitLine{CA} \parallel \drawUnitLine{ED}$,\\ -$\drawAngle{ACD} = \drawAngle{D}$ \bycref{prop:I.XXIX},\\ +$\drawAngle{ACD} = \drawAngle{D}$ \byref{prop:I.XXIX},\\ $\therefore \drawAngle{A} = \drawAngle{D}$. -И поскольку $\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{DC} : \drawUnitLine{ED}$ \bycref{\hypref},\\ -треугольники равноугольны \bycref{prop:VI.VI}. +И поскольку $\drawUnitLine{AB} : \drawUnitLine{CA} :: \drawUnitLine{DC} : \drawUnitLine{ED}$ \byref{\hypref},\\ +треугольники равноугольны \byref{prop:VI.VI}. $\therefore \drawAngle{B} = \drawAngle{DCE}$. Но $\drawAngle{A} = \drawAngle{ACD}$. -$\therefore \drawAngle{BCA} + \drawAngle{ACD} + \drawAngle{DCE} = \drawAngle{BCA} + \drawAngle{A} + \drawAngle{B} = \drawTwoRightAngles$ \bycref{prop:I.XXXII},\\ -и $\therefore$ \drawUnitLine{BC} и~\drawUnitLine{CE} лежат на одной прямой \bycref{prop:I.XIV}. +$\therefore \drawAngle{BCA} + \drawAngle{ACD} + \drawAngle{DCE} = \drawAngle{BCA} + \drawAngle{A} + \drawAngle{B} = \drawTwoRightAngles$ \byref{prop:I.XXXII},\\ +и $\therefore$ \drawUnitLine{BC} и~\drawUnitLine{CE} лежат на одной прямой \byref{prop:I.XIV}. \end{center} \qed @@ -13132,7 +13131,7 @@ \chapter*{Определения} stopGlobalRotation; } и~т. д., каждая $= \arcEF$, проведем радиусы к~концам равных дуг. -Поскольку дуги \arcBC, \arcCK, \arcKL\ и~т. д. равны, углы \drawAngle{BC}, \drawAngle{CK}, \drawAngle{KL} и~т. д. также равны \bycref{prop:III.XXVII}. $\therefore$ \drawAngle{BC,CK,KL} столько же раз кратен \drawAngle{BC}, сколько \drawFromCurrentPicture[middle][arcBL]{ +Поскольку дуги \arcBC, \arcCK, \arcKL\ и~т. д. равны, углы \drawAngle{BC}, \drawAngle{CK}, \drawAngle{KL} и~т. д. также равны \byref{prop:III.XXVII}. $\therefore$ \drawAngle{BC,CK,KL} столько же раз кратен \drawAngle{BC}, сколько \drawFromCurrentPicture[middle][arcBL]{ startGlobalRotation(180-angle(B-L)); startAutoLabeling; draw byNamedArcSeq(0)(BC,CK,KL); @@ -13147,14 +13146,14 @@ \chapter*{Определения} } кратна дуге \arcEF. \begin{center} -Тогда очевидно \bycref{prop:III.XXVII},\\ +Тогда очевидно \byref{prop:III.XXVII},\\ что если \drawAngle{BC,CK,KL} (или если $m$ раз \drawAngle{BC}) $>, =, < \drawAngle{EF,FM,MN}$ (или $n$ раз \drawAngle{EF}),\\ то \arcBL (или $m$ раз \arcBC) $>, =, < \arcEN$ (или $n$ раз \arcEF). \end{center} -$\therefore \drawAngle{BC} : \drawAngle{EF} :: \arcBC\ : \arcEF$ \bycref{def:V.V}, или углы в~центре относятся так же, как дуги, на которых они стоят, но углы на окружности вдвое меньше углов в~центре \bycref{prop:III.XX} и~относятся так же \bycref{prop:V.XV}, и, следовательно, так же, как дуги, на которых они стоят. +$\therefore \drawAngle{BC} : \drawAngle{EF} :: \arcBC\ : \arcEF$ \byref{def:V.V}, или углы в~центре относятся так же, как дуги, на которых они стоят, но углы на окружности вдвое меньше углов в~центре \byref{prop:III.XX} и~относятся так же \byref{prop:V.XV}, и, следовательно, так же, как дуги, на которых они стоят. -Очевидно, что секторы в~равных кругах и~на равных дугах равны \bycref{prop:I.IV,prop:III.XXIV,prop:III.XXVII,def:III.IX}. Значит, если выше углы заменить на секторы, получим доказательство второй части, то есть что в~равных кругах секторы относятся друг к~другу, как дуги, на которых они стоят. +Очевидно, что секторы в~равных кругах и~на равных дугах равны \byref{prop:I.IV,prop:III.XXIV,prop:III.XXVII,def:III.IX}. Значит, если выше углы заменить на секторы, получим доказательство второй части, то есть что в~равных кругах секторы относятся друг к~другу, как дуги, на которых они стоят. \qed @@ -13199,17 +13198,17 @@ \chapter*{Определения} \begin{center} Действительно, если провести $\drawSizedLine{BE} \parallel \drawSizedLine{DC}$,\\ $\begin{aligned} - \mbox{то } \drawAngle{DCB} &= \drawAngle{EBC} \mbox{ \bycref{prop:I.XXIX};}\\ - & = \drawAngle{FCD} \mbox{ \bycref{\hypref},}\\ - & = \drawAngle{E} \mbox{ \bycref{prop:I.XXIX}.} + \mbox{то } \drawAngle{DCB} &= \drawAngle{EBC} \mbox{ \byref{prop:I.XXIX};}\\ + & = \drawAngle{FCD} \mbox{ \byref{\hypref},}\\ + & = \drawAngle{E} \mbox{ \byref{prop:I.XXIX}.} \end{aligned}$ -И $\therefore \drawSizedLine{EC} = \drawSizedLine{BC}$ \bycref{prop:I.VI}\\ -и $\drawSizedLine{AE,EC} : \drawSizedLine{BC} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \bycref{prop:V.VII}. +И $\therefore \drawSizedLine{EC} = \drawSizedLine{BC}$ \byref{prop:I.VI}\\ +и $\drawSizedLine{AE,EC} : \drawSizedLine{BC} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \byref{prop:V.VII}. -Но также $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \bycref{prop:VI.II}. +Но также $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{EC}$ \byref{prop:VI.II}. -И, следовательно, $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{BC}$ \bycref{prop:V.XI}. +И, следовательно, $\drawSizedLine{AB,BD} : \drawSizedLine{BD} :: \drawSizedLine{AE,EC} : \drawSizedLine{BC}$ \byref{prop:V.XI}. \end{center} \qed @@ -13251,19 +13250,19 @@ \chapter*{Определения} Проведем \drawSizedLine{AD}, делая $\drawAngle{CAE} = \drawAngle{EAB}$,\\ тогда $\drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{BD} \times \drawSizedLine{DC} + \drawSizedLine{AD}^2$. -Около \drawLine[bottom]{CA,DC,BD,AB} опишем \drawCircle[middle][1/4]{O} \bycref{prop:IV.V},\\ +Около \drawLine[bottom]{CA,DC,BD,AB} опишем \drawCircle[middle][1/4]{O} \byref{prop:IV.V},\\ продлим \drawSizedLine{AD} до окружности и~проведем \drawSizedLine{CE}. -Поскольку $\drawAngle{CAE} = \drawAngle{EAB}$ \bycref{\hypref}\\ -и $\drawAngle{B} = \drawAngle{E}$ \bycref{prop:III.XXI},\\ -$\therefore$ \drawLine[bottom]{AD,BD,AB} и~\drawLine[middle]{CA,CE,DE,AD} равноугольны \bycref{prop:I.XXXII}. +Поскольку $\drawAngle{CAE} = \drawAngle{EAB}$ \byref{\hypref}\\ +и $\drawAngle{B} = \drawAngle{E}$ \byref{prop:III.XXI},\\ +$\therefore$ \drawLine[bottom]{AD,BD,AB} и~\drawLine[middle]{CA,CE,DE,AD} равноугольны \byref{prop:I.XXXII}. -$\therefore \drawSizedLine{AB} : \drawSizedLine{AD} :: \drawSizedLine{AD,DE} : \drawSizedLine{CA}$ \bycref{prop:VI.IV}. +$\therefore \drawSizedLine{AB} : \drawSizedLine{AD} :: \drawSizedLine{AD,DE} : \drawSizedLine{CA}$ \byref{prop:VI.IV}. -$\therefore \drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{AD} \times \drawSizedLine{AD,DE}$ \bycref{prop:VI.XVI}\\ -$= \drawSizedLine{DE} \times \drawSizedLine{AD} + \drawSizedLine{AD}^2$ \bycref{prop:II.III}. +$\therefore \drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{AD} \times \drawSizedLine{AD,DE}$ \byref{prop:VI.XVI}\\ +$= \drawSizedLine{DE} \times \drawSizedLine{AD} + \drawSizedLine{AD}^2$ \byref{prop:II.III}. -Но $\drawSizedLine{DE} \times \drawSizedLine{AD} = \drawSizedLine{BD} \times \drawSizedLine{DC}$ \bycref{prop:III.XXXV}. +Но $\drawSizedLine{DE} \times \drawSizedLine{AD} = \drawSizedLine{BD} \times \drawSizedLine{DC}$ \byref{prop:III.XXXV}. $\therefore \drawSizedLine{AB} \times \drawSizedLine{CA} = \drawSizedLine{BD} \times \drawSizedLine{DC} + \drawSizedLine{AD}^2$. \end{center} @@ -13308,14 +13307,14 @@ \chapter*{Определения} проведем $\drawUnitLine{AD} \perp \drawUnitLine{BD,DC}$;\\ тогда $\drawUnitLine{AB} \times \drawUnitLine{CA} = \drawUnitLine{AD} \times \mbox{ диаметр описанного круга}$. -Опишем \drawCircle[middle][1/4]{O} \bycref{prop:IV.V}, проведем диаметр \drawUnitLine{AE} и~проведем \drawUnitLine{CE},\\ -тогда, поскольку $\drawAngle{D} = \drawAngle{BCA,ECB}$ \bycref{\constref,prop:III.XXXI}\\ -и $\drawAngle{B} = \drawAngle{E}$ \bycref{prop:III.XXI},\\ -$\therefore$ \drawLine[bottom]{AD,BD,AB} равноуголен с~\drawLine[middle]{CA,CE,AE} \bycref{prop:VI.IV}. +Опишем \drawCircle[middle][1/4]{O} \byref{prop:IV.V}, проведем диаметр \drawUnitLine{AE} и~проведем \drawUnitLine{CE},\\ +тогда, поскольку $\drawAngle{D} = \drawAngle{BCA,ECB}$ \byref{\constref,prop:III.XXXI}\\ +и $\drawAngle{B} = \drawAngle{E}$ \byref{prop:III.XXI},\\ +$\therefore$ \drawLine[bottom]{AD,BD,AB} равноуголен с~\drawLine[middle]{CA,CE,AE} \byref{prop:VI.IV}. $\therefore \drawUnitLine{AB} : \drawUnitLine{AD} :: \drawUnitLine{AE} : \drawUnitLine{CA}$. -И $\therefore \drawUnitLine{AB} \times \drawUnitLine{CA} = \drawUnitLine{AD} \times \drawUnitLine{AE}$ \bycref{prop:VI.XVI}. +И $\therefore \drawUnitLine{AB} \times \drawUnitLine{CA} = \drawUnitLine{AD} \times \drawUnitLine{AE}$ \byref{prop:VI.XVI}. \end{center} \qed @@ -13364,20 +13363,20 @@ \chapter*{Определения} Проведем \drawUnitLine{BE,ED} и~\drawUnitLine{AC}, тогда $\drawUnitLine{BE,ED} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD} + \drawUnitLine{DA} \times \drawUnitLine{BC}$. -Сделаем $\drawAngle{EAB} = \drawAngle{DAC}$ \bycref{prop:I.XXIII},\\ -$\therefore \drawAngle{EAB,CAE} = \drawAngle{CAE,DAC}$ и~$\drawAngle{BCA} = \drawAngle{D}$ \bycref{prop:III.XXI}. +Сделаем $\drawAngle{EAB} = \drawAngle{DAC}$ \byref{prop:I.XXIII},\\ +$\therefore \drawAngle{EAB,CAE} = \drawAngle{CAE,DAC}$ и~$\drawAngle{BCA} = \drawAngle{D}$ \byref{prop:III.XXI}. -$\therefore \drawUnitLine[0.8cm]{DA} : \drawUnitLine[0.8cm]{ED} :: \drawUnitLine[0.8cm]{AC} : \drawUnitLine[0.8cm]{BC}$ \bycref{prop:VI.IV},\\ -и $\therefore \drawUnitLine[0.75cm]{ED} \times \drawUnitLine[0.75cm]{AC} = \drawUnitLine[0.75cm]{DA} \times \drawUnitLine[0.75cm]{BC}$ \bycref{prop:VI.XVI}. +$\therefore \drawUnitLine[0.8cm]{DA} : \drawUnitLine[0.8cm]{ED} :: \drawUnitLine[0.8cm]{AC} : \drawUnitLine[0.8cm]{BC}$ \byref{prop:VI.IV},\\ +и $\therefore \drawUnitLine[0.75cm]{ED} \times \drawUnitLine[0.75cm]{AC} = \drawUnitLine[0.75cm]{DA} \times \drawUnitLine[0.75cm]{BC}$ \byref{prop:VI.XVI}. Теперь, поскольку\\ -$\drawAngle{EAB} = \drawAngle{DAC}$ \bycref{\constref} и $\drawAngle{B} = \drawAngle{ACD}$ \bycref{prop:III.XXI},\\ -$\therefore \drawUnitLine{AB} : \drawUnitLine{BE} :: \drawUnitLine{AC} : \drawUnitLine{CD}$ \bycref{prop:VI.IV},\\ -и $\therefore \drawUnitLine[0.75cm]{BE} \times \drawUnitLine[0.75cm]{AC} = \drawUnitLine[0.75cm]{AB} \times \drawUnitLine[0.75cm]{CD}$ \bycref{prop:VI.XVI}. +$\drawAngle{EAB} = \drawAngle{DAC}$ \byref{\constref} и $\drawAngle{B} = \drawAngle{ACD}$ \byref{prop:III.XXI},\\ +$\therefore \drawUnitLine{AB} : \drawUnitLine{BE} :: \drawUnitLine{AC} : \drawUnitLine{CD}$ \byref{prop:VI.IV},\\ +и $\therefore \drawUnitLine[0.75cm]{BE} \times \drawUnitLine[0.75cm]{AC} = \drawUnitLine[0.75cm]{AB} \times \drawUnitLine[0.75cm]{CD}$ \byref{prop:VI.XVI}. Но, как написано выше, $\drawUnitLine{ED} \times \drawUnitLine{AC} = \drawUnitLine{DA} \times \drawUnitLine{BC}$. -$\therefore \drawUnitLine{BE,ED} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD} + \drawUnitLine{DA} \times \drawUnitLine{BC}$ \bycref{prop:II.I}. +$\therefore \drawUnitLine{BE,ED} \times \drawUnitLine{AC} = \drawUnitLine{AB} \times \drawUnitLine{CD} + \drawUnitLine{DA} \times \drawUnitLine{BC}$ \byref{prop:II.I}. \end{center} \qed diff --git a/byrnebook.cls b/byrnebook.cls index 404c1b6..041eb9f 100644 --- a/byrnebook.cls +++ b/byrnebook.cls @@ -31,8 +31,9 @@ \usepackage{extramarks} \usepackage{csquotes} \usepackage{fontspec} -\usepackage{cleveref} - +\usepackage{zref-clever} +\zcsetup{abbrev} +\usepackage[main=\byrnebook@booklanguage]{babel} \pagestyle{fancy} \fancyhead[LE,RO]{\thepage} @@ -583,7 +584,7 @@ % Other % -\newcommand\bycref[1]{(\cref{#1})} +\newcommand\byref[1]{(\zcref{#1})} \newcommand\charspacing[2]{\addfontfeature{LetterSpace=#1} #2 \addfontfeature{LetterSpace=0}} \emergencystretch 3em @@ -606,17 +607,286 @@ \def\bookString{Book} -\crefname{axiom}{\inaxstr}{\inaxstr} -\crefname{definition}{\indefstr}{\indefstr} -\crefname{definitionAZ}{\indefstr}{\indefstr} -\crefname{postulate}{\inpoststr}{\inpoststr} -\crefname{proposition}{\inpropstr}{\inpropstr} -\crefname{propositionAZ}{\inpropstr}{\inpropstr} -\crefname{construction}{}{} -\crefname{hypothesis}{}{} +\zcLanguageSetup{english}{ + type = axiom , + Name-sg = Axiom , + name-sg = axiom , + Name-pl = Axioms , + name-pl = axioms , + Name-sg-ab = Ax. , + name-sg-ab = ax. , + Name-pl-ab = Ax. , + name-pl-ab = ax. , + type = definition , + Name-sg = Definition , + name-sg = definition , + Name-pl = Definitions , + name-pl = definitions , + Name-sg-ab = Def. , + name-sg-ab = def. , + Name-pl-ab = Def. , + name-pl-ab = def. , + type = postulate , + Name-sg = Postulate , + name-sg = postulate , + Name-pl = Postulate , + name-pl = postulate , + Name-sg-ab = Post. , + name-sg-ab = post. , + Name-pl-ab = Post. , + name-pl-ab = post. , + type = proposition , + Name-sg = Proposition , + name-sg = proposition , + Name-pl = Propositions , + name-pl = propositions , + Name-sg-ab = Prop. , + name-sg-ab = prop. , + Name-pl-ab = Prop. , + name-pl-ab = prop. , + } + +\zcLanguageSetup{russian}{ + type = axiom , + gender = f , + case = n , + Name-sg = Аксиома , + name-sg = аксиома , + Name-pl = Аксиомы , + name-pl = аксиомы , + Name-sg-ab = Акс. , + name-sg-ab = акс. , + Name-pl-ab =Акс. , + name-pl-ab = акс. , + case = a , + Name-sg = Аксиому , + name-sg = аксиому , + Name-pl = Аксиомы , + name-pl = аксиомы , + Name-sg-ab = Акс. , + name-sg-ab = акс. , + Name-pl-ab =Акс. , + name-pl-ab = акс. , + case = g , + Name-sg = Аксиомы , + name-sg = аксиомы , + Name-pl = Аксиом , + name-pl = аксиом , + Name-sg-ab = Акс. , + name-sg-ab = акс. , + Name-pl-ab =Акс. , + name-pl-ab = акс. , + case = d , + Name-sg = Аксиоме , + name-sg = аксиоме , + Name-pl = Аксиомам , + name-pl = аксиомам , + Name-sg-ab = Акс. , + name-sg-ab = акс. , + Name-pl-ab =Акс. , + name-pl-ab = акс. , + case = i , + Name-sg = Аксиомой , + name-sg = аксиомой , + Name-pl = Аксиомами , + name-pl = аксиомами , + Name-sg-ab = Акс. , + name-sg-ab = акс. , + Name-pl-ab =Акс. , + name-pl-ab = акс. , + case = p , + Name-sg = Аксиоме , + name-sg = аксиоме , + Name-pl = Аксиомах , + name-pl = аксиомах , + Name-sg-ab = Акс. , + name-sg-ab = акс. , + Name-pl-ab =Акс. , + name-pl-ab = акс. , + type = definition , + gender = f , + case = n , + Name-sg = Определение , + name-sg = определение , + Name-pl = Определения , + name-pl = определения , + Name-sg-ab = Опр. , + name-sg-ab = опр. , + Name-pl-ab =Опр. , + name-pl-ab = опр. , + case = a , + Name-sg = Определение , + name-sg = определение , + Name-pl = Определения , + name-pl = определения , + Name-sg-ab = Опр. , + name-sg-ab = опр. , + Name-pl-ab =Опр. , + name-pl-ab = опр. , + case = g , + Name-sg = Определения , + name-sg = определения , + Name-pl = Определений , + name-pl = определений , + Name-sg-ab = Опр. , + name-sg-ab = опр. , + Name-pl-ab =Опр. , + name-pl-ab = опр. , + case = d , + Name-sg = Определению , + name-sg = определению , + Name-pl = Определениям , + name-pl = определениям , + Name-sg-ab = Опр. , + name-sg-ab = опр. , + Name-pl-ab =Опр. , + name-pl-ab = опр. , + case = i , + Name-sg = Определением , + name-sg = определением , + Name-pl = Определениями , + name-pl = определениями , + Name-sg-ab = Опр. , + name-sg-ab = опр. , + Name-pl-ab =Опр. , + name-pl-ab = опр. , + case = p , + Name-sg = Определении , + name-sg = определении , + Name-pl = Определениях , + name-pl = определениях , + Name-sg-ab = Опр. , + name-sg-ab = опр. , + Name-pl-ab =Опр. , + name-pl-ab = опр. , + type = postulate , + gender = m , + case = n , + Name-sg = Постулат , + name-sg = постулат , + Name-pl = Постулаты , + name-pl = постулаты , + Name-sg-ab = Пост. , + name-sg-ab = пост. , + Name-pl-ab =Пост. , + name-pl-ab = пост. , + case = a , + Name-sg = Постулат , + name-sg = постулат , + Name-pl = Постулаты , + name-pl = постулаты , + Name-sg-ab = Пост. , + name-sg-ab = пост. , + Name-pl-ab =Пост. , + name-pl-ab = пост. , + case = g , + Name-sg = Постулата , + name-sg = постулата , + Name-pl = Постулатов , + name-pl = постулатов , + Name-sg-ab = Пост. , + name-sg-ab = пост. , + Name-pl-ab =Пост. , + name-pl-ab = пост. , + case = d , + Name-sg = Постулату , + name-sg = постулату , + Name-pl = Постулатам , + name-pl = постулатам , + Name-sg-ab = Пост. , + name-sg-ab = пост. , + Name-pl-ab =Пост. , + name-pl-ab = пост. , + case = i , + Name-sg = Постулатом , + name-sg = постулатом , + Name-pl = Постулатами , + name-pl = постулатами , + Name-sg-ab = Пост. , + name-sg-ab = пост. , + Name-pl-ab =Пост. , + name-pl-ab = пост. , + case = p , + Name-sg = Постулате , + name-sg = постулате , + Name-pl = Постулатах , + name-pl = постулатах , + Name-sg-ab = Пост. , + name-sg-ab = пост. , + Name-pl-ab =Пост. , + name-pl-ab = пост. , + type = proposition , + gender = n , + case = n , + Name-sg = Предложение , + name-sg = предложение , + Name-pl = Предложения , + name-pl = предложения , + Name-sg-ab = Пр. , + name-sg-ab = пр. , + Name-pl-ab = Пр. , + name-pl-ab = пр. , + case = a , + Name-sg = Предложение , + name-sg = предложение , + Name-pl = Предложения , + name-pl = предложения , + Name-sg-ab = Пр. , + name-sg-ab = пр. , + Name-pl-ab = Пр. , + name-pl-ab = пр. , + case = g , + Name-sg = Предложения , + name-sg = предложения , + Name-pl = Предложений , + name-pl = предложений , + Name-sg-ab = Пр. , + name-sg-ab = пр. , + Name-pl-ab = Пр. , + name-pl-ab = пр. , + case = d , + Name-sg = Предложению , + name-sg = предложению , + Name-pl = Предложениям , + name-pl = предложениям , + Name-sg-ab = Пр. , + name-sg-ab = пр. , + Name-pl-ab = Пр. , + name-pl-ab = пр. , + case = i , + Name-sg = Предложением , + name-sg = предложением , + Name-pl = Предложениями , + name-pl = предложениями , + Name-sg-ab = Пр. , + name-sg-ab = пр. , + Name-pl-ab = Пр. , + name-pl-ab = пр. , + case = p , + Name-sg = Предложении , + name-sg = предложении , + Name-pl = Предложениях , + name-pl = предложениях , + Name-sg-ab = Пр. , + name-sg-ab = пр. , + Name-pl-ab = Пр. , + name-pl-ab = пр. , + } + + +\zcRefTypeSetup{hypothesis}{ + Name-sg = , + name-sg = , + Name-pl = , + name-pl = , +} -\creflabelformat{construction}{#2#1#3} -\creflabelformat{hypothesis}{#2#1#3} +\zcRefTypeSetup{construction}{ + Name-sg = , + name-sg = , + Name-pl = , + name-pl = , +} \AtBeginDocument{ \@ifpackagewith{babel}{russian}{% @@ -629,14 +899,6 @@ \def\indefstr{опр.}% \def\inaxstr{акс.}% \def\bookString{Книга}% - \crefname{axiom}{\inaxstr}{\inaxstr}% - \crefname{definition}{\indefstr}{\indefstr}% - \crefname{postulate}{\inpoststr}{\inpoststr}% - \crefname{proposition}{\inpropstr}{\inpropstr}% - \crefname{construction}{}{}% - \crefname{hypothesis}{}{}% - \creflabelformat{construction}{#2#1#3}% - \creflabelformat{hypothesis}{#2#1#3} % }{} }