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-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-drawing.tex72
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-examples.tex204
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-intersec.tex83
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.pdfbin998875 -> 0 bytes
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.tex10
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-others.tex55
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-presentation.tex23
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-show.tex2
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex54
-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/tkz-euclide.pdfbin998875 -> 1015168 bytes
10 files changed, 346 insertions, 157 deletions
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-drawing.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-drawing.tex
index e4d08474d0..181285901c 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-drawing.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-drawing.tex
@@ -255,25 +255,61 @@ This is of course equivalent to \tkzcname{draw (A)--(B);}. You can also use the
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{Adding dimensions with option \tkzname{dim} new code from Muzimuzhi Z}
+This code comes from an answer to this question on tex.stackexchange.com
+(change-color-and-style-of-dimension-lines-in-tkz-euclide )
+You can use now two styles : |dim style| and |dim fence style|. You have several ways to use them.
+I'll let you look at the examples to see what you can do with these styles.
+
+\begin{verbatim}
+ \tikzset{dim style/.append style={dashed}} % append if you want to keep precedent style.
+ or
+ \begin{scope}[ dim style/.append style={orange},
+ dim fence style/.style={dashed}]
+\end{verbatim}
+
+
+\begin{tkzexample}[latex=7cm]
+\begin{tikzpicture}[scale=.75]
+ \tkzDefPoints{0/3/A, 1/-3/B}
+ \tkzDrawPoints(A,B)
+ \tkzDrawSegment[dim={\(l_0\),1cm,right=2mm},
+ dim style/.append style={red,
+ dash pattern={on 2pt off 2pt}}](A,B)
+ \tkzDrawSegment[dim={\(l_1\),2cm,right=2mm},
+ dim style/.append style={blue}](A,B)
+ \begin{scope}[ dim style/.style={orange},
+ dim fence style/.style={dashed}]
+ \tkzDrawSegment[dim={\(l_2\),3cm,right=2mm}](A,B)
+ \tkzDrawSegment[dim={\(l_3\),-2cm,right=2mm}](A,B)
+ \end{scope}
+ \tkzLabelPoints[left](A,B)
+\end{tikzpicture}
+\end{tkzexample}
+
+
\subsubsection{Adding dimensions with option \tkzname{dim} partI}
-\begin{tkzexample}[vbox,small]
-\begin{tikzpicture}[scale=4]
- \pgfkeys{/pgf/number format/.cd,fixed,precision=2}
- % Define the first two points
- \tkzDefPoint(0,0){A}
- \tkzDefPoint(3,0){B}
- \tkzDefPoint(1,1){C}
- % Draw the triangle and the points
- \tkzDrawPolygon(A,B,C)
- \tkzDrawPoints(A,B,C)
- % Label the sides
- \tkzCalcLength(A,B)\tkzGetLength{ABl}
- \tkzCalcLength(B,C)\tkzGetLength{BCl}
- \tkzCalcLength(A,C)\tkzGetLength{ACl}
- % add dim
- \tkzDrawSegment[dim={\pgfmathprintnumber\BCl,6pt,transform shape}](C,B)
- \tkzDrawSegment[dim={\pgfmathprintnumber\ACl,6pt,transform shape}](A,C)
- \tkzDrawSegment[dim={\pgfmathprintnumber\ABl,-6pt,transform shape}](A,B)
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}[scale=2]
+\pgfkeys{/pgf/number format/.cd,fixed,precision=2}
+\tkzDefPoint(0,0){A}
+\tkzDefPoint(3.07,0){B}
+\tkzInterCC[R](A,2.37)(B,1.82)
+\tkzGetPoints{C}{C'}
+\tkzDrawCircle[in](A,B,C) \tkzGetPoint{G}
+\tkzGetLength{rIn}
+\tkzDrawPolygon(A,B,C)
+\tkzDrawPoints(A,B,C)
+\tkzCalcLength(A,B)\tkzGetLength{ABl}
+\tkzCalcLength(B,C)\tkzGetLength{BCl}
+\tkzCalcLength(A,C)\tkzGetLength{ACl}
+\begin{scope}[dim style/.style={dashed,sloped,teal}]
+ \tkzDrawSegment[dim={\pgfmathprintnumber\BCl,6pt,
+ text=red}](C,B)
+ \tkzDrawSegment[dim={\pgfmathprintnumber\ACl,6pt,}](A,C)
+ \tkzDrawSegment[dim={\pgfmathprintnumber\ABl,-6pt,}](A,B)
+\end{scope}
+\tkzLabelPoints(A,B) \tkzLabelPoints[above](C)
\end{tikzpicture}
\end{tkzexample}
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-examples.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-examples.tex
index 79b340b5db..60ecb19965 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-examples.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-examples.tex
@@ -41,7 +41,7 @@
\tkzDefPointWith[orthogonal,K=-1](B,A)
\tkzDrawLine[add = .5 and .5](B,tkzPointResult)
\tkzInterLC[R](B,tkzPointResult)(A,8)
- \tkzGetPoints{C}{J}
+ \tkzGetPoints{J}{C}
\tkzDrawSegment(A,B)
\tkzDrawPoints(A,B,C)
\tkzCompass(A,C)
@@ -67,7 +67,7 @@ The figure below shows a semicircle, with diameter $AB$. A tangent line is drawn
\tkzDefPoint(0,0){A}\tkzDefPoint(6,0){D}
\tkzDefPoint(8,0){B}\tkzDefPoint(4,0){I}
\tkzDefLine[orthogonal=through D](A,D)
- \tkzInterLC[R](D,tkzPointResult)(I,4) \tkzGetFirstPoint{C}
+ \tkzInterLC[R](D,tkzPointResult)(I,4) \tkzGetSecondPoint{C}
\tkzDefLine[orthogonal=through C](I,C) \tkzGetPoint{c}
\tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b}
\tkzInterLL(C,c)(B,b) \tkzGetPoint{T}
@@ -201,7 +201,7 @@ The figure below shows a semicircle, with diameter $AB$. A tangent line is drawn
\tkzInit[xmin=-1,xmax=11,ymin=-4,ymax=7]
\tkzClip
\tkzDefPoints{0/0/A,8/0/B}
- \tkzDefTriangle[pythagore](A,B)
+ \tkzDefTriangle[pythagore,swap](A,B)
\tkzGetPoint{C}
\tkzDrawPolygon[fill=green!5](A,B,C)
\tkzDefMidPoint(C,A) \tkzGetPoint{I}
@@ -348,7 +348,7 @@ At what angle does he see these two vertices?}
\tkzDefPointWith[orthogonal](I,Y) \tkzGetPoint{i}
\tkzDrawLines[add = 2 and 1,color=orange](I,i)
\tkzInterLL(I,i)(A,B) \tkzGetPoint{Z}
- \tkzInterLC(I,i)(O,B) \tkzGetSecondPoint{M}
+ \tkzInterLC(I,i)(O,B) \tkzGetFirstPoint{M}
\tkzDefPointWith[orthogonal](B,Z) \tkzGetPoint{b}
\tkzDrawCircle(O,B)
\tkzDrawLines[add = 0 and 2,color=orange](B,b)
@@ -379,7 +379,7 @@ The three altitudes of a triangle intersect at the same H-point.
\tkzDefPoint(5,6){A}
\tkzDefMidPoint(C,B) \tkzGetPoint{I}
\tkzInterLC(A,C)(I,B)
- \tkzGetSecondPoint{B'}
+ \tkzGetFirstPoint{B'}
\tkzInterLC(A,B)(I,B)
\tkzGetFirstPoint{C'}
\tkzInterLL(B,B')(C,C') \tkzGetPoint{H}
@@ -403,8 +403,8 @@ The three altitudes of a triangle intersect at the same H-point.
\tkzDefPoint(5,6){C}
\tkzDefMidPoint(A,B)\tkzGetPoint{O}
\tkzDefPointBy[projection=onto A--B](C) \tkzGetPoint{P}
-\tkzInterLC(C,A)(O,A)
-\tkzGetSecondPoint{M}
+\tkzInterLC[common=A](C,A)(O,A)
+\tkzGetFirstPoint{M}
\tkzInterLC(C,B)(O,A)
\tkzGetFirstPoint{N}
\tkzInterLL(B,M)(A,N)\tkzGetPoint{I}
@@ -919,8 +919,8 @@ Another solution
\tkzInterLC[R](A,B)(A,1) \tkzGetPoints{D}{a2}
\tkzDefMidPoint(D,E) \tkzGetPoint{I}
\tkzDrawCircle[orange](I,D)
-\tkzInterLC(X,H)(I,D) \tkzGetPoints{M'}{M}
-\tkzInterLC(M,D)(A,D) \tkzGetPoints{P'}{P}
+\tkzInterLC(X,H)(I,D) \tkzGetPoints{M}{M'}
+\tkzInterLC(M,D)(A,D) \tkzGetPoints{P}{P'}
\tkzInterLC(M,E)(B,E) \tkzGetPoints{Q}{Q'}
\tkzInterLL(P,Q)(A,B) \tkzGetPoint{O}
\tkzDrawSegments[orange](A,P I,M B,Q)
@@ -931,28 +931,28 @@ Another solution
\end{tikzpicture}
\begin{tkzexample}[code only,small]
- \begin{tikzpicture}
- \tkzDefPoints{0/0/A,4/2/B,2/3/K}
- \tkzDrawCircle[R](A,1)\tkzDrawCircle[R](B,3)
- \tkzInterCC[R](A,1)(K,3) \tkzGetPoints{a}{a'}
- \tkzInterCC[R](B,3)(K,3) \tkzGetPoints{b}{b'}
- \tkzInterLL(a,a')(b,b') \tkzGetPoint{X}
- \tkzDefPointBy[projection= onto A--B](X) \tkzGetPoint{H}
- \tkzGetPoint{C}
- \tkzInterLC[R](A,B)(B,3) \tkzGetPoints{b1}{E}
- \tkzInterLC[R](A,B)(A,1) \tkzGetPoints{D}{a2}
- \tkzDefMidPoint(D,E) \tkzGetPoint{I}
- \tkzDrawCircle[orange](I,D)
- \tkzInterLC(X,H)(I,D) \tkzGetPoints{M'}{M}
- \tkzInterLC(M,D)(A,D) \tkzGetPoints{P'}{P}
- \tkzInterLC(M,E)(B,E) \tkzGetPoints{Q}{Q'}
- \tkzInterLL(P,Q)(A,B) \tkzGetPoint{O}
- \tkzDrawSegments[orange](A,P I,M B,Q)
- \tkzDrawPoints(A,B,D,E,M,I,O,P,Q,X,H)
- \tkzDrawLines(O,E M,D M,E O,Q)
- \tkzDrawLine[add= 3 and 4,orange](X,H)
- \tkzLabelPoints(A,B,D,E,M,I,O,P,Q,X,H)
- \end{tikzpicture}
+\begin{tikzpicture}
+\tkzDefPoints{0/0/A,4/2/B,2/3/K}
+\tkzDrawCircle[R](A,1)\tkzDrawCircle[R](B,3)
+\tkzInterCC[R](A,1)(K,3) \tkzGetPoints{a}{a'}
+\tkzInterCC[R](B,3)(K,3) \tkzGetPoints{b}{b'}
+\tkzInterLL(a,a')(b,b') \tkzGetPoint{X}
+\tkzDefPointBy[projection= onto A--B](X) \tkzGetPoint{H}
+\tkzGetPoint{C}
+\tkzInterLC[R](A,B)(B,3) \tkzGetPoints{b1}{E}
+\tkzInterLC[R](A,B)(A,1) \tkzGetPoints{D}{a2}
+\tkzDefMidPoint(D,E) \tkzGetPoint{I}
+\tkzDrawCircle[orange](I,D)
+\tkzInterLC(X,H)(I,D) \tkzGetPoints{M}{M'}
+\tkzInterLC(M,D)(A,D) \tkzGetPoints{P}{P'}
+\tkzInterLC(M,E)(B,E) \tkzGetPoints{Q}{Q'}
+\tkzInterLL(P,Q)(A,B) \tkzGetPoint{O}
+\tkzDrawSegments[orange](A,P I,M B,Q)
+\tkzDrawPoints(A,B,D,E,M,I,O,P,Q,X,H)
+\tkzDrawLines(O,E M,D M,E O,Q)
+\tkzDrawLine[add= 3 and 4,orange](X,H)
+\tkzLabelPoints(A,B,D,E,M,I,O,P,Q,X,H)
+\end{tikzpicture}
\end{tkzexample}
@@ -1105,10 +1105,10 @@ Since the triangles $AO_2C$ and $AO_1E$ are isosceles the angles at the base are
\tkzInterCC(O,A)(M_0,B) \tkzGetFirstPoint{D}
\tkzInterLL(O_1,E)(O_2,F) \tkzGetPoint{O_3}
\tkzDefCircle[circum](E,F,B) \tkzGetPoint{0_4}
- \tkzInterLC(A,D)(O_1,A) \tkzGetSecondPoint{I}
- \tkzInterLC(C,D)(O_2,B) \tkzGetFirstPoint{K}
- \tkzInterLC(A,D)(O_3,D) \tkzGetFirstPoint{G}
- \tkzInterLC(C,D)(O_3,D) \tkzGetSecondPoint{H}
+ \tkzInterLC(A,D)(O_1,A) \tkzGetFirstPoint{I}
+ \tkzInterLC(C,D)(O_2,B) \tkzGetSecondPoint{K}
+ \tkzInterLC[common=D](A,D)(O_3,D) \tkzGetFirstPoint{G}
+ \tkzInterLC[common=D](C,D)(O_3,D) \tkzGetFirstPoint{H}
\tkzInterLL(C,G)(B,K) \tkzGetPoint{M}
\tkzInterLL(A,H)(B,I) \tkzGetPoint{L}
\tkzInterLL(L,G)(A,C) \tkzGetPoint{N}
@@ -1188,9 +1188,17 @@ The last example is very complex and it is to show you all that we can do with \
% !TEX TS-program = lualatex
\documentclass{standalone}
\usepackage{tkz-euclide}
+\tkzSetUpColors[background=white,text=black]
+\tkzSetUpCompass[color=orange, line width=.4pt,delta=10]
+\tkzSetUpArc[color=gray,line width=.4pt]
+\tkzSetUpPoint[size=2,color=teal]
+\tkzSetUpLine[line width=.4pt,color=teal]
+\tkzSetUpStyle[orange]{new}
+\tikzset{every picture/.style={line width=.4pt}}
+
\begin{document}
-\begin{tikzpicture}[scale=1]
+\begin{tikzpicture}[scale=.75]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
% we need some special points if the triangle, tkz-euclide.sty knows about them
@@ -1218,22 +1226,20 @@ The last example is very complex and it is to show you all that we can do with \
\tkzInterLL(Zc,Ya)(A,B) \tkzGetPoint{A'}
\tkzDefPointBy[reflection= over Ja--Jc](C')\tkzGetPoint{Ab}
\tkzDefPointBy[reflection= over Ja--Jc](A')\tkzGetPoint{Cb}
-
% Now we can get the center of THE CIRCLE : Q
% BUT we need to find the radius or a point on the circle
\tkzInterLL(K,O)(N,Sp) \tkzGetPoint{Q}
-\tkzInterLC(A,B)(Q,Cb) \tkzGetSecondPoint{Ba}
-\tkzInterLC(A,C)(Q,Cb) \tkzGetPoints{Ca}{Ac}
-\tkzInterLC(B,C')(Q,Cb) \tkzGetSecondPoint{Bc}
+\tkzInterLC(A,B)(Q,Cb) \tkzGetFirstPoint{Ba}
+\tkzInterLC(A,C)(Q,Cb) \tkzGetPoints{Ac}{Ca}
+\tkzInterLC(B,C')(Q,Cb) \tkzGetFirstPoint{Bc}
\tkzInterLC(Q,Ja)(Q,Cb) \tkzGetSecondPoint{F'a}
\tkzInterLC(Q,Jc)(Q,Cb) \tkzGetSecondPoint{F'c}
\tkzInterLC(Q,Jb)(Q,Cb) \tkzGetSecondPoint{F'b}
\tkzInterLC(Sp,F'a)(Ja,Za) \tkzGetFirstPoint{Fa}
\tkzInterLC(Sp,F'b)(Jb,Yb) \tkzGetFirstPoint{Fb}
-\tkzInterLC(Sp,F'c)(Jc,Yc) \tkzGetSecondPoint{Fc}
-\tkzInterLC(Mc,Sp)(Q,Cb) \tkzGetSecondPoint{A''}
+\tkzInterLC(Sp,F'c)(Jc,Yc) \tkzGetFirstPoint{Fc}
+\tkzInterLC(Mc,Sp)(Q,Cb) \tkzGetFirstPoint{A''}
\tkzDefLine[parallel=through A''](N,Mc) \tkzGetPoint{q}
-
% Calculations are done, now you can draw, mark and label
\tkzDrawPolygon(A,B,C)
\tkzDrawCircle(Q,Bc)%
@@ -1262,66 +1268,70 @@ The last example is very complex and it is to show you all that we can do with \
\tkzMarkRightAngles(Jc,Zc,A Ja,Xa,B Jb,Yb,C)
\tkzDrawSegments[green,dashed](A,F'a B,F'b C,F'c)
\end{tikzpicture}
-
\end{document}
\end{tkzexample}
\subsubsection*{The result}
%
\begin{tikzpicture}[scale=.6]
- \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
- \tkzDefTriangleCenter[euler](A,B,C) \tkzGetPoint{N}
- \tkzDefTriangleCenter[circum](A,B,C) \tkzGetPoint{O}
- \tkzDefTriangleCenter[lemoine](A,B,C) \tkzGetPoint{K}
- \tkzDefTriangleCenter[ortho](A,B,C) \tkzGetPoint{H}
- \tkzDefSpcTriangle[excentral,name=J](A,B,C){a,b,c}
- \tkzDefSpcTriangle[centroid,name=M](A,B,C){a,b,c}
- \tkzDefCircle[in](Ma,Mb,Mc) \tkzGetPoint{Sp} % Sp Spieker center
- \tkzDefProjExcenter[name=J](A,B,C)(a,b,c){Y,Z,X}
- \tkzDefLine[parallel=through Za](A,B) \tkzGetPoint{Xc}
- \tkzInterLL(Za,Xc)(C,B) \tkzGetPoint{C'}
- \tkzDefLine[parallel=through Zc](B,C) \tkzGetPoint{Ya}
- \tkzInterLL(Zc,Ya)(A,B) \tkzGetPoint{A'}
- \tkzDefPointBy[reflection= over Ja--Jc](C')\tkzGetPoint{Ab}
- \tkzDefPointBy[reflection= over Ja--Jc](A')\tkzGetPoint{Cb}
- \tkzInterLL(K,O)(N,Sp) \tkzGetPoint{Q}
- \tkzInterLC(A,B)(Q,Cb) \tkzGetSecondPoint{Ba}
- \tkzInterLC(A,C)(Q,Cb) \tkzGetPoints{Ca}{Ac}
- \tkzInterLC(B,C')(Q,Cb) \tkzGetSecondPoint{Bc}
- \tkzInterLC(Q,Ja)(Q,Cb) \tkzGetSecondPoint{F'a}
- \tkzInterLC(Q,Jc)(Q,Cb) \tkzGetSecondPoint{F'c}
- \tkzInterLC(Q,Jb)(Q,Cb) \tkzGetSecondPoint{F'b}
- \tkzInterLC(Sp,F'a)(Ja,Za) \tkzGetFirstPoint{Fa}
- \tkzInterLC(Sp,F'b)(Jb,Yb) \tkzGetFirstPoint{Fb}
- \tkzInterLC(Sp,F'c)(Jc,Yc) \tkzGetSecondPoint{Fc}
- \tkzInterLC(Mc,Sp)(Q,Cb) \tkzGetSecondPoint{A''}
- \tkzDefLine[parallel=through A''](N,Mc) \tkzGetPoint{q}
- \tkzDrawPolygon(A,B,C)
- \tkzDrawCircle(Q,Bc)%
- \tkzDrawCircle[euler,lightgray](A,B,C)
- \tkzDrawCircles[ex](A,B,C B,C,A C,A,B)
- \tkzDrawSegments[dashed](A,A' C,C' A',Zc Za,C' B,Cb B,Ab A,Ca C,Ac Ja,Xa Jb,Yb Jc,Zc)
- \begin{scope}
- \tkzClipCircle(Q,Cb)
- \tkzDrawLine[add=5 and 12,orange](K,O)
- \tkzDrawLine[add=12 and 28,red!50!black](N,Sp)
- \end{scope}
- \tkzDrawSegments[color=green!50!black](Mc,N Mc,A'' A'',Q)
- \tkzDrawSegments[color=red,dashed](Ac,Ab Ca,Cb Ba,Bc Ja,Jc A',Cb C',Ab)
- \tkzDrawSegments[color=red](Cb,Ab Bc,Ac Ba,Ca A',C')
- \tkzMarkSegments[color=red,mark=|](Cb,Ab Bc,Ac Ba,Ca)
- \tkzMarkRightAngles(Jc,Zc,A Ja,Xa,B Jb,Yb,C)
- \tkzDrawSegments[green,dashed](A,F'a B,F'b C,F'c)
- \tkzDrawPoints(A,B,C,K,Ja,Jb,Jc,Q,N,O,Sp,Mc,Xa,Xb,Yb,Yc,Za,Zc,
- A',C',A'',Ab,Cb,Bc,Ca,Ac,Ba,Fa,Fb,Fc,F'a,F'b,F'c)
- \tkzLabelPoints(Ja,Jb,Jc,Q,Xa,Xb,Za,Zc,Ab,Cb,
- Bc, Ca, Ac, Ba, F'b)
- \tkzLabelPoints[above](O, K, F'a, Fa, A'')
- \tkzLabelPoints[below](B, F'c, Yc, N, Sp, Fc, Mc)
- \tkzLabelPoints[left](A', C', Fb)
- \tkzLabelPoints[right](C)
- \tkzLabelPoints[below right](A)
- \tkzLabelPoints[above right](Yb)
+\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
+\tkzDefTriangleCenter[euler](A,B,C) \tkzGetPoint{N}
+\tkzDefTriangleCenter[circum](A,B,C) \tkzGetPoint{O}
+\tkzDefTriangleCenter[lemoine](A,B,C) \tkzGetPoint{K}
+\tkzDefTriangleCenter[ortho](A,B,C) \tkzGetPoint{H}
+\tkzDefSpcTriangle[excentral,name=J](A,B,C){a,b,c}
+\tkzDefSpcTriangle[centroid,name=M](A,B,C){a,b,c}
+\tkzDefCircle[in](Ma,Mb,Mc) \tkzGetPoint{Sp}
+
+\tkzDefProjExcenter[name=J](A,B,C)(a,b,c){Y,Z,X}
+\tkzDefLine[parallel=through Za](A,B) \tkzGetPoint{Xc}
+\tkzInterLL(Za,Xc)(C,B) \tkzGetPoint{C'}
+\tkzDefLine[parallel=through Zc](B,C) \tkzGetPoint{Ya}
+\tkzInterLL(Zc,Ya)(A,B) \tkzGetPoint{A'}
+\tkzDefPointBy[reflection= over Ja--Jc](C')\tkzGetPoint{Ab}
+\tkzDefPointBy[reflection= over Ja--Jc](A')\tkzGetPoint{Cb}
+
+\tkzInterLL(K,O)(N,Sp) \tkzGetPoint{Q}
+\tkzInterLC(A,B)(Q,Cb) \tkzGetFirstPoint{Ba}
+\tkzInterLC(A,C)(Q,Cb) \tkzGetPoints{Ac}{Ca}
+\tkzInterLC(B,C')(Q,Cb) \tkzGetFirstPoint{Bc}
+\tkzInterLC(Q,Ja)(Q,Cb) \tkzGetSecondPoint{F'a}
+\tkzInterLC(Q,Jc)(Q,Cb) \tkzGetSecondPoint{F'c}
+\tkzInterLC(Q,Jb)(Q,Cb) \tkzGetSecondPoint{F'b}
+\tkzInterLC(Sp,F'a)(Ja,Za) \tkzGetFirstPoint{Fa}
+\tkzInterLC(Sp,F'b)(Jb,Yb) \tkzGetFirstPoint{Fb}
+\tkzInterLC(Sp,F'c)(Jc,Yc) \tkzGetFirstPoint{Fc}
+\tkzInterLC(Mc,Sp)(Q,Cb) \tkzGetFirstPoint{A''}
+\tkzDefLine[parallel=through A''](N,Mc) \tkzGetPoint{q}
+\tkzDrawPolygon(A,B,C)
+\tkzDrawCircle(Q,Bc)%
+\tkzDrawCircle[euler,lightgray](A,B,C)
+\tkzDrawCircles[ex](A,B,C B,C,A C,A,B)
+\tkzDrawSegments[dashed](A,A' C,C' A',Zc Za,C' B,Cb B,Ab A,Ca C,Ac
+ Ja,Xa Jb,Yb Jc,Zc)
+
+\begin{scope}
+ \tkzClipCircle(Q,Cb)
+ \tkzDrawLine[add=5 and 12,orange](K,O)
+ \tkzDrawLine[add=12 and 28,red!50!black](N,Sp)
+\end{scope}
+
+\tkzDrawPoints(A,B,C,K,Ja,Jb,Jc,Q,N,O,Sp,Mc,Xa,Xb,Yb,Yc,Za,Zc)
+\tkzDrawPoints(A',C',A'',Ab,Cb,Bc,Ca,Ac,Ba,Fa,Fb,Fc,F'a,F'b,F'c)
+\tkzLabelPoints(Ja,Jb,Jc,Q,Xa,Xb,Za,Zc,Ab,Cb,Bc,Ca,Ac,Ba,F'b)
+\tkzLabelPoints[above](O,K,F'a,Fa,A'')
+\tkzLabelPoints[below](B,F'c,Yc,N,Sp,Fc,Mc)
+\tkzLabelPoints[left](A',C',Fb)
+\tkzLabelPoints[right](C)
+\tkzLabelPoints[below right](A)
+\tkzLabelPoints[above right](Yb)
+\tkzDrawSegments[color=green!50!black](Mc,N Mc,A'' A'',Q)
+\tkzDrawSegments[color=red,dashed](Ac,Ab Ca,Cb Ba,Bc Ja,Jc A',Cb C',Ab)
+\tkzDrawSegments[color=red](Cb,Ab Bc,Ac Ba,Ca A',C')
+\tkzMarkSegments[color=red,mark=|](Cb,Ab Bc,Ac Ba,Ca)
+\tkzMarkRightAngles(Jc,Zc,A Ja,Xa,B Jb,Yb,C)
+\tkzDrawSegments[green,dashed](A,F'a B,F'b C,F'c)
\end{tikzpicture}
+
\endinput \ No newline at end of file
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-intersec.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-intersec.tex
index 2fb014f30c..a8676a2854 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-intersec.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-intersec.tex
@@ -45,6 +45,7 @@ options & default & definition \\
\TOline{N} {N} { (O,C) determines the circle}
\TOline{R} {N} { (O, 1 ) unit 1 cm}
\TOline{with nodes}{N} { (O,C,D) CD is a radius}
+\TOline{common} {} { common = pt if pt is common point}
\bottomrule
\end{tabular}
@@ -72,6 +73,41 @@ In the following example, the drawing of the circle uses two points and the inte
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{Line-circle intersection with common point}
+\begin{tkzexample}[latex=7cm,small]
+ \begin{tikzpicture}[scale=.5]
+ \tkzDefPoints{0/0/O,5/1/A,2/2/B}
+ \tkzInterLC[common=A](B,A)(O,A)\tkzGetFirstPoint{C}
+ \tkzDrawPoints(O,A,B)
+ \tkzDrawCircle(O,A)
+ \tkzDrawLine(A,C)
+ \tkzDrawPoint(C)
+ \tkzLabelPoints(A,B,C)
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+\subsubsection{Line-circle intersection order of points}
+The idea is to compare the angles formed with the first defining point of the line, a resultant point and the center of the circle. The first point is the one that corresponds to the smallest angle.
+
+As you can see $\widehat{BCO} < \widehat{BEO} $
+
+\begin{tkzexample}[latex=7cm,small]
+ \begin{tikzpicture}[scale=.5]
+ \tkzDefPoints{0/0/O,5/1/A,2/2/B,3/1/D}
+ \tkzInterLC[common=A](B,D)(O,A) \tkzGetPoints{C}{E}
+ \tkzDrawPoints(O,A,B,D)
+ \tkzDrawCircle(O,A)
+ \tkzDrawLine(E,C)
+ \tkzDrawSegments[dashed](B,O O,C)
+ \tkzMarkAngle[->,size=1.5](B,C,O)
+ \tkzDrawSegments[dashed](O,E)
+ \tkzMarkAngle[->,size=1.5](B,E,O)
+ \tkzDrawPoints(C,E)
+ \tkzLabelPoints(O,A,B,C,D,E)
+ \end{tikzpicture}
+\end{tkzexample}
+
\subsubsection{Line-circle intersection in Sangaku}
\begin{tkzexample}[vbox,small]
@@ -95,15 +131,15 @@ In the following example, the drawing of the circle uses two points and the inte
\tkzDefPoint["$B$" below right](\ORadius,0){B}
\tkzDefPoint["$O_2$" below left](\OORadius-\ORadius,0){O2}
\tkzDefLine[mediator](A,B) \tkzGetPoints{mr}{ml}
- \tkzInterLC[R](D,mr)(O,\ORadius) \tkzGetPoints{C}{E}
+ \tkzInterLC[R](D,mr)(O,\ORadius) \tkzGetPoints{E}{C}
\tkzDefLine[orthogonal=through A](X,A) \tkzGetPoint{pr}
\ifdim\XA pt < 0 pt\relax
- \tkzInterLC[R](A,pr)(O,\OOOORadius) \tkzGetPoints{O4}{O3}
+ \tkzInterLC[R](A,pr)(O,\OOOORadius) \tkzGetPoints{O3}{O4}
\else
\ifdim\XA pt = 0pt\relax
- \tkzInterLC[R](A,pr)(O,\OOOORadius) \tkzGetPoints{O4}{O3}
- \else
\tkzInterLC[R](A,pr)(O,\OOOORadius) \tkzGetPoints{O3}{O4}
+ \else
+ \tkzInterLC[R](A,pr)(O,\OOOORadius) \tkzGetPoints{O4}{O3}
\fi
\fi
\tkzDefPointBy[projection=onto A--C](O3) \tkzGetPoint{H}
@@ -252,7 +288,42 @@ This macro defines the intersection point(s) $I$ and $J$ of the two center circl
It is also possible to use directly \tkzcname{tkzInterCCN} and \tkzcname{tkzInterCCR}.
\end{NewMacroBox}
-\subsubsection{Construction of an equilateral triangle}
+\subsubsection{circle-circle intersection with common point.}
+
+\begin{tkzexample}[latex=7cm,small]
+ \begin{tikzpicture}[scale=.5]
+ \tkzDefPoints{0/0/O,5/-1/A,2/2/B}
+ \tkzDrawPoints(O,A,B)
+ \tkzDrawCircles(O,B A,B)
+ \tkzInterCC[common=B](O,B)(A,B)\tkzGetFirstPoint{C}
+ \tkzDrawPoint(C)
+ \tkzLabelPoints(O,A,B,C)
+ \end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{circle-circle intersection order of points.}
+The idea is to compare the angles formed with the first center, a resultant point and the center of the second circle. The first point is the one that corresponds to the smallest angle.
+
+As you can see $\widehat{ODB} < \widehat{OBE} $
+
+\begin{tkzexample}[latex=7cm,small]
+ \begin{tikzpicture}[scale=.5]
+ \tkzDefPoints{0/0/O,5/-1/A,2/2/B,2/-1/C}
+ \tkzDrawPoints(O,A,B)
+ \tkzDrawCircles(O,A B,C)
+ \tkzInterCC(O,A)(B,C)\tkzGetPoints{D}{E}
+ \tkzDrawPoints(C,D,E)
+ \tkzLabelPoints(O,A,B,C,D,E)
+ \tkzDrawSegments[dashed](D,O D,B)
+ \tkzMarkAngle[->,size=1.5](O,D,B)
+ \tkzDrawSegments[dashed](E,O E,B)
+ \tkzMarkAngle[->,size=1.5](O,E,B)
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+
+\subsubsection{Construction of an equilateral triangle.}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[trim left=-1cm,scale=.5]
\tkzDefPoint(1,1){A}
@@ -269,7 +340,7 @@ It is also possible to use directly \tkzcname{tkzInterCCN} and \tkzcname{tkzInte
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Example a mediator}
+\subsubsection{Example a mediator.}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoint(0,0){A}
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.pdf b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.pdf
deleted file mode 100644
index 91d7ecba05..0000000000
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.pdf
+++ /dev/null
Binary files differ
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.tex
index 6b0f8f1c17..e02856ecd3 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.tex
@@ -18,15 +18,15 @@
index = totoc,
twoside,
headings = small,
- cadre
+ %cadre
]{tkz-doc}
%\usepackage{etoc}
\gdef\tkznameofpack{tkz-euclide}
-\gdef\tkzversionofpack{4.00}
-\gdef\tkzdateofpack{2022/01/04}
+\gdef\tkzversionofpack{4.03}
+\gdef\tkzdateofpack{2022/01/20}
\gdef\tkznameofdoc{doc-tkz-euclide}
-\gdef\tkzversionofdoc{4.00}
-\gdef\tkzdateofdoc{2022/01/04}
+\gdef\tkzversionofdoc{4.03}
+\gdef\tkzdateofdoc{2022/01/20}
\gdef\tkzauthorofpack{Alain Matthes}
\gdef\tkzadressofauthor{}
\gdef\tkznamecollection{AlterMundus}
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-others.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-others.tex
index 30af7dd713..47886c814d 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-others.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-others.tex
@@ -1,4 +1,3 @@
-
\section{Different authors}
\subsection{Code from Andrew Swan}
@@ -224,7 +223,7 @@ Below the figure used to illustrate the proof of the “Gou gu theorem.” (cod
\tkzCalcLength(A',B) \tkzGetLength{lB}
\pgfmathparse{\lA-\lB}
\tkzInterLC[R](A,A')(A',\pgfmathresult)
- \tkzGetFirstPoint{D'}
+ \tkzGetSecondPoint{D'}
\tkzDefSquare(D',A')\tkzGetPoints{B'}{C'}
\tkzDefLine[orthogonal=through D](D,D')
\tkzGetPoint{d}
@@ -279,7 +278,6 @@ Advanced Mathematics, Drawing geometry pictures.
\end{tikzpicture}
\begin{tikzpicture}
-
\tkzDefPoint(0,0){A} \tkzDefPoint(4,1){B}
\tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{D}
\tkzInterLC(A,B)(B,A) \tkzGetPoints{F}{E}
@@ -288,13 +286,13 @@ Advanced Mathematics, Drawing geometry pictures.
\tkzCompasss[color=red, very thick](A,C B,C A,D B,D)
\begin{scope}
- \tkzSetUpArc[fill=blue!10,thick,delta=0]
- \tkzDrawArc(A,B)(C)
- \tkzDrawArc(B,C)(A)
- \tkzDrawArc(C,A)(B)
+ \tkzSetUpArc[thick,delta=0]
+ \tkzDrawArc[fill=blue!10](A,B)(C)
+ \tkzDrawArc[fill=blue!10](B,C)(A)
+ \tkzDrawArc[fill=blue!10](C,A)(B)
\end{scope}
-
- \tkzMarkAngles[fill=yellow,opacity=0.5](D,A,E A,E,D)
+ \tkzMarkAngles(D,A,E A,E,D)
+ \tkzFillAngles[fill=yellow,opacity=0.5](D,A,E A,E,D)
\tkzMarkRightAngle[size=0.65,fill=red!20,opacity=0.2](A,D,E)
\tkzLabelAngle[pos=0.7](D,A,E){$\alpha$}
@@ -316,4 +314,43 @@ Advanced Mathematics, Drawing geometry pictures.
\end{tikzpicture}
+\begin{tkzexample}[code only,small]
+ \begin{tikzpicture}
+ \tkzDefPoint(0,0){A} \tkzDefPoint(4,1){B}
+ \tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{D}
+ \tkzInterLC(A,B)(B,A) \tkzGetPoints{F}{E}
+ \tkzDrawCircles[dashed](A,B B,A)
+ \tkzDrawPolygons(A,B,C A,E,D)
+ \tkzCompasss[color=red, very thick](A,C B,C A,D B,D)
+ \begin{scope}
+ \tkzSetUpArc[thick,delta=0]
+ \tkzDrawArc[fill=blue!10](A,B)(C)
+ \tkzDrawArc[fill=blue!10](B,C)(A)
+ \tkzDrawArc[fill=blue!10](C,A)(B)
+ \end{scope}
+ \tkzMarkAngles(D,A,E A,E,D)
+ \tkzFillAngles[fill=yellow,opacity=0.5](D,A,E A,E,D)
+ \tkzMarkRightAngle[size=0.65,fill=red!20,opacity=0.2](A,D,E)
+ \tkzLabelAngle[pos=0.7](D,A,E){$\alpha$}
+ \tkzLabelAngle[pos=0.8](A,E,D){$\beta$}
+ \tkzLabelAngle[pos=0.5,xshift=-1.4mm](A,D,D){$90^\circ$}
+ \begin{scope}[font=\small]
+ \tkzLabelSegment[below=0.6cm,align=center](A,B){Reuleaux\\triangle}
+ \tkzLabelSegment[above right,sloped](A,E){hypotenuse}
+ \tkzLabelSegment[below,sloped](D,E){opposite}
+ \tkzLabelSegment[below,sloped](A,D){adjacent}
+ \tkzLabelSegment[below right=4cm](A,E){Thales circle}
+ \end{scope}
+ \tkzLabelPoints[below left](A)
+ \tkzLabelPoints(B,D)
+ \tkzLabelPoint[above](C){$C$}
+ \tkzLabelPoints(E)
+ \tkzDrawPoints(A,...,E)
+
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+
+
\endinput \ No newline at end of file
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-presentation.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-presentation.tex
index 826c521f80..56af228c28 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-presentation.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-presentation.tex
@@ -92,7 +92,7 @@ We want to draw a circle around the points $A$ and $B$ whose radius is given by
|\tkzDrawCircles(A,B B,A)|
}
-The intersection of the circles
+The intersection of the circles $\mathcal{D}$ and $\mathcal{E}$
\medskip
@@ -158,7 +158,7 @@ How to draw points :
Explanation
-In the first part, we need to find the midpoint of the straight line $AB$. With TikZ we can use the calc library
+In the first part, we need to find the midpoint of the straight line $AB$. With \TIKZ\ we can use the calc library
\medskip
\hspace*{1cm}\vbox{\orange |\coordinate [label=left:$A$] (A) at (0,0);|\\
@@ -193,7 +193,7 @@ We can draw the triangle at the end of the picture with
\hspace*{1cm}\vbox{\red |\tkzDrawPolygon{A,B,C}|}
\medskip
-We know how to draw the circle around $B$ through $C$ and how to place the points $E$ and $F$
+We know how to draw the circle $\mathcal{H}$ around $B$ through $C$ and how to place the points $E$ and $F$
\medskip
\hspace*{1cm}\vbox{\orange
@@ -209,8 +209,8 @@ We know how to draw the circle around $B$ through $C$ and how to place the point
\medskip
We can place the points $E$ and $F$ at the end of the picture. We don't need them now.
-Intersecting a Line and a Circle : here we search the intersection of the circle around B through C and the line DB.
-The infinite straight line DB intercepts the circle but with TikZ we need to extend the lines DB and that can be done using partway calculations. We get the point F and BF or DF intercepts the circle
+Intersecting a Line and a Circle : here we search the intersection of the circle around $B$ through $C$ and the line $DB$.
+The infinite straight line $DB$ intercepts the circle but with \TIKZ\ we need to extend the lines $DB$ and that can be done using partway calculations. We get the point $F$ and $BF$ or $DF$ intercepts the circle
\medskip
\hspace*{1cm}\vbox{\orange| \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {}; | \\
@@ -219,14 +219,17 @@ The infinite straight line DB intercepts the circle but with TikZ we need to ext
|\fill[red,opacity=.5] (G) circle (2pt);|} \\
\medskip
-Like the intersection of two circles, it's easy to find the intersection of a line and a circle with \pkg{elements}. We don't need $F$
+Like the intersection of two circles, it's easy to find the intersection of a line and a circle with \pkg{tkz-euclide}. We don't need $F$
\medskip
\hspace*{1cm}\vbox{\red | \tkzInterLC(B,D)(B,C)\tkzGetFirstPoint{G}|}
\medskip
-there are no more difficulties. Here the final code with some simplications.
+There are no more difficulties. Here the final code with some simplications.
+Nous tracons le cercle $\mathcal{K}$ de centre $D$ et passant par $G$. Il coupe la droite $AD$ au point $L$. $AL = BC$.
+\hspace*{1cm}\vbox{\red | \tkzDrawCircle(D,G)|}
+\hspace*{1cm}\vbox{\red | \tkzInterLC(D,A)(D,G)\tkzGetSecondPoint{L}|}
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[scale=2]
@@ -240,15 +243,15 @@ there are no more difficulties. Here the final code with some simplications.
\tkzDrawLines[add=0 and 2](D,A D,B)
\tkzDrawSegment(A,B)
\tkzDrawSegments[red](A,L B,C)
- \tkzDrawPoints[red](D,L)
+ \tkzDrawPoints[red](D,L,G)
\tkzDrawPoints[fill=gray](A,B,C)
\tkzLabelPoints[left,red](A)
\tkzLabelPoints[below right,red](L)
- \tkzLabelCircle[above left=6pt](B,G)(180){$H$}
+ \tkzLabelCircle[above left=6pt](B,G)(180){$\mathcal{H}$}
\tkzLabelPoints[above left](D,G)
\tkzLabelPoints[above,red](C)
\tkzLabelPoints[right,red](B)
- \tkzLabelCircle[above left=6pt](D,G)(180){$K$}
+ \tkzLabelCircle[above left=6pt](D,G)(180){$\mathcal{K}$}
\end{tikzpicture}
\end{tkzexample}
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-show.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-show.tex
index 3a35b4ba2b..01bcc1bce7 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-show.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-show.tex
@@ -147,7 +147,7 @@ You'll find this figure again, but without the construction features.
\tkzDefMidPoint(A,B) \tkzGetPoint{O}
\tkzDefPointBy[projection=onto A--B](I)
\tkzGetPoint{J}
- \tkzInterLC(I,A)(O,A) \tkzGetPoints{M'}{M}
+ \tkzInterLC(I,A)(O,A) \tkzGetPoints{M}{M'}
\tkzInterLC(I,B)(O,A) \tkzGetPoints{N}{N'}
\tkzDrawSemiCircle[diameter](A,B)
\tkzDrawSegments(I,A I,B A,B B,M A,N)
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
index 5becdf0e2f..990862afff 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
@@ -28,13 +28,17 @@ options & default & definition \\
\midrule
\TOline{two angles= \#1 and \#2}{no defaut}{triangle knowing two angles}
\TOline{equilateral} {equilateral}{equilateral triangle }
+\TOline{half} {equilateral}{B rectangle $AB=2BC$ $AC$ hypothenuse }
\TOline{isosceles right} {equilateral}{isosceles right triangle }
\TOline{pythagore}{equilateral}{proportional to the pythagorean triangle 3-4-5}
+\TOline{pythagoras}{equilateral}{same as above}
+\TOline{egyptian}{equilateral}{same as above}
\TOline{school} {equilateral}{angles of 30, 60 and 90 degrees }
\TOline{gold}{equilateral}{angles of 72, 72 and 36 degrees, $A$ is the apex}
\TOline{euclid} {equilateral}{same as above but $[AB]$ is the base}
\TOline{golden} {equilateral}{B rectangle and $AB/AC = \Phi$}
\TOline{cheops} {equilateral}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.}
+\TOline{swap} {false}{gives the symmetric point with respect to $AB$}
\bottomrule
\end{tabular}
@@ -42,6 +46,22 @@ options & default & definition \\
\tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use.
\end{NewMacroBox}
+\subsubsection{Option \tkzname{equilateral}}
+\begin{tkzexample}[latex=7 cm,small]
+\begin{tikzpicture}
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(4,0){B}
+ \tkzDefTriangle[equilateral](A,B)
+ \tkzGetPoint{C}
+ \tkzDrawPolygons(A,B,C)
+ \tkzDefTriangle[equilateral](B,A)
+ \tkzGetPoint{D}
+ \tkzDrawPolygon(B,A,D)
+ \tkzMarkSegments[mark=s|](A,B B,C A,C A,D B,D)
+\end{tikzpicture}
+\end{tkzexample}
+
+
\subsubsection{Option \tkzname{two angles}}
\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}
@@ -95,6 +115,23 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5.
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{Option \tkzname{pythagore} and \tkzname{swap}}
+This triangle has sides whose lengths are proportional to 3, 4 and 5.
+
+\begin{tkzexample}[latex=6 cm,small]
+\begin{tikzpicture}
+ \tkzDefPoints{0/0/A,4/0/B}
+ \tkzDefTriangle[pythagore,swap](A,B)
+ \tkzGetPoint{C}
+ \tkzDrawSegments(A,B)
+ \tkzDrawSegments[new](A,C B,C)
+ \tkzMarkRightAngles(A,B,C)
+ \tkzLabelPoint[above,new](C){$C$}
+ \tkzDrawPoints[new](C)
+ \tkzDrawPoints(A,B)
+ \tkzLabelPoints(A,B)
+\end{tikzpicture}
+\end{tkzexample}
\subsubsection{Option \tkzname{golden}}
\begin{tkzexample}[latex=6 cm,small]
@@ -111,22 +148,17 @@ This triangle has sides whose lengths are proportional to 3, 4 and 5.
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Option \tkzname{equilateral} and \tkzname{isosceles right}}
+\subsubsection{Option \tkzname{isosceles right}}
\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(4,0){B}
- \tkzDefTriangle[equilateral](A,B)
- \tkzGetPoint{C}
\tkzDefTriangle[isosceles right](A,B)
- \tkzGetPoint{E}
- \tkzDrawPolygons(A,B,C A,B,E)
- \tkzDefTriangle[equilateral](B,A)
- \tkzGetPoint{D}
- \tkzDrawPolygon(B,A,D)
- \tkzMarkRightAngles(B,E,A)
- \tkzDrawPoints(A,B,C,D,E)
- \tkzLabelPoints(A,B,C,D,E)
+ \tkzGetPoint{C}
+ \tkzDrawPolygons(A,B,C)
+ \tkzDrawPoints(A,B,C)
+ \tkzMarkRightAngles(A,C,B)
+ \tkzLabelPoints(A,B,C)
\end{tikzpicture}
\end{tkzexample}
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/tkz-euclide.pdf b/macros/latex/contrib/tkz/tkz-euclide/doc/tkz-euclide.pdf
index 91d7ecba05..f00c25c789 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/tkz-euclide.pdf
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/tkz-euclide.pdf
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