diff options
Diffstat (limited to 'macros/latex/contrib/tkz/tkz-euclide/doc')
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 Binary files differdeleted file mode 100644 index 91d7ecba05..0000000000 --- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-main.pdf +++ /dev/null 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 Binary files differindex 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 |