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-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/README.md29
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-drawing.tex72
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-examples.tex204
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-intersec.tex83
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.pdfbin998875 -> 0 bytes
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.tex10
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-others.tex55
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex23
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-show.tex2
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex54
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/tkz-euclide.pdfbin998875 -> 1015168 bytes
11 files changed, 374 insertions, 158 deletions
diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/README.md b/Master/texmf-dist/doc/latex/tkz-euclide/README.md
index 54f230b2e05..cda513e6f61 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/README.md
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/README.md
@@ -1,6 +1,6 @@
# tkz-euclide — for euclidean geometry
-Release 4.00 b 2022/01/04
+Release 4.03 b 2022/01/19
## Description
@@ -73,6 +73,33 @@ The new version of `tkz-euclide` is *not* fully compatible with the version
3.06 but the differences are minor.
## History
+
+- 4.03 Adaptation of the code and documentation to the changes of the macros for the intersections.
+- 4.02
+ Major changes for the macros concerning the intersection of a line and a circle or two circles. If one point of the intersection is known then you can use the "common" option and indicate what the common point is. The second point is given in tkzFirstPointResult.
+ In other cases, for the intersection of two circles the determined points form angles with the centers of the circles. One of the angles measures less than 180 degrees and the other more than 180. The smaller one determines tkzFirstPointResult.
+ For the intersection of a line and a circle, the method is the same except that the angle is formed by a point on the line and the center of the circle.
+
+- 4.01
+ \tkzDefOrthogonalCircle was defined twice so I deleted the version in tkz-obj-eu-circles-by
+
+ In the tkz-obj-eu-draw-lines.tex new code for add dim from muzimuzhi Z.
+ The code comes from an answer on the site tex.stackexchange.com
+
+ In the file tkz-obj-eu-draw-triangles.tex added options
+ pythagoras and egyptian equivalent to pythagore
+ euclid equivalent to euclide
+ two one equivalent to half
+
+ Added option "swap" useful with golden, gold, school, half, pythagoras
+
+ In the file tkz-obj-eu-circles correction of bug in \tkzDefOrthoThroughCircle : \tkz@@CalcLength has been replaced by \tkz@@CalcLengthcm
+
+ Addition of the macro \tkzDefGoldenRatio in tkz-obj-eu-points-spc. It allows to split a segment with a ratio equal to the golden ratio
+
+ Minor corrections of the documentation. New examples about option "dim"
+
+
- 4.00 correction of bugs, tkz-euclide no longer depends on tkz-base. The unit is "cm".
The bounding box is controlled. The documentation has been restructured according to the rule:
set, calculate, draw, mark and fill, label.
diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-drawing.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-drawing.tex
index e4d08474d03..181285901ca 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-drawing.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-examples.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-examples.tex
index 79b340b5db6..60ecb199658 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-examples.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-intersec.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-intersec.tex
index 2fb014f30c1..a8676a28544 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-intersec.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.pdf b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.pdf
deleted file mode 100644
index 91d7ecba05a..00000000000
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.pdf
+++ /dev/null
Binary files differ
diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.tex
index 6b0f8f1c177..e02856ecd37 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-main.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-others.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-others.tex
index 30af7dd713d..47886c814de 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-others.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex
index 826c521f805..56af228c28e 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-show.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-show.tex
index 3a35b4ba2ba..01bcc1bce7f 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-show.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex
index 5becdf0e2f1..990862afff2 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-triangles.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/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/Master/texmf-dist/doc/latex/tkz-euclide/tkz-euclide.pdf b/Master/texmf-dist/doc/latex/tkz-euclide/tkz-euclide.pdf
index 91d7ecba05a..f00c25c789c 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/tkz-euclide.pdf
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/tkz-euclide.pdf
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