automatic angle naming

README.md

@@ -1 +1,4 @@
 MetaPost + ConTeXt rendition of Oliver Byrne's "The first six books of the Elements of Euclid"
+
+The book itself (byrne_context.tex) is licensed under CC-BY-SA 4.0.
+MetaPost library (byrne.mp) and lettrines generator (lettrines.mp) are licensed under GPLv3 or later.

byrne_context.tex

@@ -117,10 +117,10 @@
 {\tfa github.com/jemmybutton}
 \vskip 0.25\baselineskip
 
-{\tfb 2017 ed.\,0.1}
+{\tfb 2017 ed.\,0.1a}
 \vskip \baselineskip
 
-\symbol[cc][cc-by-sa-nc] 
+\symbol[cc][cc] \symbol[cc][by] \symbol[cc][sa] 
 
 \startnarrower
 \setuplocalinterlinespace[line=2ex]
@@ -206,8 +206,8 @@
 \startitemize[m,joinedup,nowhite]
 \item $\drawAngle{A} + \drawAngle{B} + \drawAngle{C} = 2 \drawAngle{B} = \drawTwoRightAngles$. \\ That is, the red angle added to the yellow angle added to the blue angle, equal twice the yellow angle, equal two right angles.
 \item $\drawAngle{A} + \drawAngle{C} = \drawAngle{B}$. \\ Or in words, the red angle added to the blue angle, equal the yellow angle.
-\item $\drawAngle{B} < \drawAngle{A} \mbox{ or } < \drawAngle{C}$. \\ The yellow angle is greater that either the red or blue angle.
-\item $\drawAngle{A} \mbox{ or } \drawAngle{C} > \drawAngle{B}$. \\ Either the red or blue angle is less that the yellow angle.
+\item $\drawAngle{B} > \drawAngle{A} \mbox{ or } > \drawAngle{C}$. \\ The yellow angle is greater that either the red or blue angle.
+\item $\drawAngle{A} \mbox{ or } \drawAngle{C} < \drawAngle{B}$. \\ Either the red or blue angle is less that the yellow angle.
 \item $\drawAngle{B} \mbox{ minus } \drawAngle{C} = \drawAngle{A}$. \\ In other terms, the yellow angle made less be the blue angle equal red angle.
 \item $\drawUnitLine{BC}^2 = \drawUnitLine{AB}^2 + \drawUnitLine{CA}^2$. \\ That is, the square of the yellow line is equal to the sum of the squares of the blue and red lines.
 \stopitemize
@@ -1336,13 +1336,15 @@
 C := B xscaled -1;
 D := 9/5[A,B];
 E := 9/5[A,C];
-draw byAngleWithName(B, A, C, black, 0)(A);
-draw byAngleWithName(A, B, C, byblue, 0)(B);
-draw byAngleWithName(B, C, A, byblue, 0)(C);
+draw byAngle(B, A, C, black, 0);
+draw byAngle(A, B, C, byblue, 0);
+draw byAngle(B, C, A, byblue, 0);
 draw byAngle(C, B, E, byyellow, 0);
 draw byAngle(D, C, B, byyellow, 0);
-draw byAngleWithName(B, D, C, byred, 0)(D);
-draw byAngleWithName(C, E, B, byred, 0)(E);
+draw byAngle(B, D, C, byred, 0);
+draw byAngle(C, E, B, byred, 0);
+byAngleDefine(E, B, D, black, 1);
+byAngleDefine(D, C, E, black, 1);
 byLineDefine(B, D, byyellow, 0, 0);
 byLineDefine(C, E, byyellow, 0, 0);
 byLineDefine(B, E, byblue, 0, 0);
@@ -1351,8 +1353,6 @@
 byLineDefine(A, C, byred, 0, 0);
 byLineDefine(B, C, black, 0, 0);
 draw byNamedLineSeq(0)(BE,noLine,BC,noLine,CD,BD,AB,AC,CE);
-byAngleDefine(E, B, D, white, 0);
-byAngleDefine(D, C, E, white, 0);
 }
 \drawCurrentPictureInMargin
 \problemNP{I}{n}{any isosceles triangle \drawLine[bottom]{AB,AC,BC} if the equal sides be produced, the external angles at the base are equal, and the internal angles at the base are also equal.}
@@ -1364,26 +1364,23 @@
 
 Then in
 \drawFromCurrentPicture{
-draw byNamedAngle(A);
+draw byNamedAngle(BAC);
 draw byNamedLineSeq(0)(AB,AC,CE,BE);
 }
 and
 \drawFromCurrentPicture{
-draw byNamedAngle(A);
+draw byNamedAngle(BAC);
 draw byNamedLineSeq(0)(AB,AC,CD,BD);
 }\\
 we have $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$ (const.),\\ 
-\drawAngle{A} common to both,\\ 
+\drawAngle{BAC} common to both,\\ 
 and $\drawUnitLine{AB} = \drawUnitLine{AC}$ (hyp.)\\ 
-$\therefore \drawAngle{C,DCB} = \drawAngle{B,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ and $\drawAngle{E} = \drawAngle{D}$ \inprop[prop:I.IV].
+$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ and $\drawAngle{CEB} = \drawAngle{BDC}$ \inprop[prop:I.IV].
 
-Again in
-\drawLine{BC,CE,BE}
-and
-\drawLine{BC,CD,BD}\\
-we have $\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{E} = \drawAngle{D}$ and $\drawUnitLine{BE} = \drawUnitLine{CD}$,\\ 
+Again in \drawLine{BC,CE,BE} and \drawLine{BC,CD,BD}\\
+we have $\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{CEB} = \drawAngle{BDC}$ and $\drawUnitLine{BE} = \drawUnitLine{CD}$,\\ 
 $\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ and $\drawAngle{DCB} = \drawAngle{CBE}$ \inprop[prop:I.IV]\\ 
-but $\drawAngle{C,DCB} = \drawAngle{B,CBE}$, $\therefore \drawAngle{C} = \drawAngle{B}$.
+but $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$.
 \stopCenterAlign
 
 \qed