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diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-example.tex b/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-example.tex new file mode 100644 index 00000000000..cba1eecd49b --- /dev/null +++ b/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-example.tex @@ -0,0 +1,347 @@ +%!TEX root = /Users/ego/Boulot/TKZ/tkz-euclide/doc_fr/TKZdoc-euclide-main.tex + +\section{Gallery : Some examples} + +Some examples with explanations in english. +%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + +\subsection{White on Black} +This example shows how to get a segment with a length equal at $\sqrt{a}$ from a segment of length $a$, only with a rule and a compass. + + +\begin{center} +\begin{tkzexample}[] + \tikzset{background rectangle/.style={fill=black}} +\begin{tikzpicture}[show background rectangle] + \tkzInit[ymin=-1.5,ymax=7,xmin=-1,xmax=+11] + \tkzClip + \tkzDefPoint(0,0){O} + \tkzDefPoint(1,0){I} + \tkzDefPoint(10,0){A} + \tkzDefPointWith[orthogonal](I,A) \tkzGetPoint{H} + \tkzDefMidPoint(O,A) \tkzGetPoint{M} + \tkzInterLC(I,H)(M,A)\tkzGetPoints{C}{B} + \tkzDrawSegments[color=white,line width=1pt](I,H O,A) + \tkzDrawPoints[color=white](O,I,A,B,M) + \tkzMarkRightAngle[color=white,line width=1pt](A,I,B) + \tkzDrawArc[color=white,line width=1pt,style=dashed](M,A)(O) + \tkzLabelSegment[white,right=1ex,pos=.5](I,B){$\sqrt{a}$} + \tkzLabelSegment[white,below=1ex,pos=.5](O,I){$1$} + \tkzLabelSegment[pos=.6,white,below=1ex](I,A){$a$} +\end{tikzpicture} +\end{tkzexample} +\end{center} + +\vfill\newpage +%<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + +\subsection{ Square root of the integers } +How to get $1$, $\sqrt{2}$, $\sqrt{3}$ with a rule and a compass. +\begin{center} +\begin{tkzexample}[] +\begin{tikzpicture}[scale=1.75] + \tkzInit[xmin=-3,xmax=4,ymin=-2,ymax=4] + \tkzGrid + \tkzDefPoint(0,0){O} + \tkzDefPoint(1,0){a0} + \newcounter{tkzcounter} + \setcounter{tkzcounter}{0} + \newcounter{density} + \setcounter{density}{20} + \foreach \i in {0,...,15}{% + \pgfmathsetcounter{density}{\thedensity+2} + \setcounter{density}{\thedensity} + \stepcounter{tkzcounter} + \tkzDefPointWith[orthogonal normed](a\i,O) + \tkzGetPoint{a\thetkzcounter} + \tkzDrawPolySeg[color=Maroon!\thedensity,% + fill=Maroon!\thedensity,opacity=.5](a\i,a\thetkzcounter,O)} + \end{tikzpicture} +\end{tkzexample} +\end{center} + +%<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + \vfill\newpage +%<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> +% +\subsection{How to construct the tangent lines from a point to a circle with a rule and a compass.} +\begin{center} +\begin{tkzexample}[] + \begin{tikzpicture} + \tkzPoint(0,0){O} + \tkzPoint(9,2){P} + \tkzDefMidPoint(O,P) \tkzGetPoint{I} + \tkzDrawCircle[R](O,4cm) + \tkzDrawCircle[diameter](O,P) + \tkzCalcLength(I,P) \tkzGetLength{dIP} + \tkzInterCC[R](O,4cm)(I,\dIP pt)\tkzGetPoints{Q1}{Q2} + \tkzDrawPoint[color=red](Q1) + \tkzDrawPoint[color=red](Q2) + \tkzDrawLine(P,Q1) + \tkzDrawLine(P,Q2) + \tkzDrawSegments(O,Q1 O,Q2) + \tkzDrawLine(P,O) +\end{tikzpicture} +\end{tkzexample} +\end{center} +% +% %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + \vfill\newpage +%<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + +\subsection{Circle and tangent} +We have a point A $(8,2)$, a circle with center A and radius=3cm and a line + $\delta$ $y=4$. The line intercepts the circle at B. We want to draw the tangent at the circle in B. + +\begin{center} +\begin{tkzexample}[] +\begin{tikzpicture} + \tkzInit[xmax=14,ymin=-2,ymax=6] + \tkzDrawX[noticks,label=$(d)$] + \tkzPoint[pos=above right](8,2){A}; + \tkzPoint[color=red,pos=above right](0,0){O}; + \tkzDrawCircle[R,color=blue,line width=.8pt](A,3 cm) + \tkzHLine[color=red,style=dashed]{4} + \tkzText[above](12,4){$\delta$} + \FPeval\alphaR{arcsin(2/3)}% on a les bonnes valeurs + \FPeval\xB{8-3*cos(\alphaR)} + \tkzPoint[pos=above left](\xB,4){B}; + \tkzDrawSegment[line width=1pt](A,B) + \tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b} + \tkzDefPoint(1,0){i} + \tkzInterLL(B,b)(O,i) \tkzGetPoint{B'} + \tkzDrawPoint(B') + \tkzDrawLine(B,B') + \end{tikzpicture} +\end{tkzexample} +\end{center} + + \vfill\newpage +%<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + +\subsection{About right triangle} + +We have a segment $[AB]$ and we want to determine a point $C$ such as $AC=8 cm$ and $ABC$ is a right triangle in $B$. + +\begin{center} +\begin{tkzexample}[] +\begin{tikzpicture} + \tkzInit + \tkzClip + \tkzPoint[pos=left](2,1){A} + \tkzPoint(6,4){B} + \tkzDrawSegment(A,B) + \tkzDrawPoint[color=red](A) + \tkzDrawPoint[color=red](B) + \tkzDefPointWith[orthogonal,K=-1](B,A) + \tkzDrawLine[add = .5 and .5](B,tkzPointResult) + \tkzInterLC[R](B,tkzPointResult)(A,8 cm) \tkzGetPoints{C}{J} + \tkzDrawPoint[color=red](C) + \tkzCompass(A,C) + \tkzMarkRightAngle(A,B,C) + \tkzDrawLine[color=gray,style=dashed](A,C) +\end{tikzpicture} +\end{tkzexample} +\end{center} + + %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + \vfill\newpage %<–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––> + +\subsection{Archimedes} + +This is an ancient problem proved by the great Greek mathematician Archimedes . +The figure below shows a semicircle, with diameter $AB$. A tangent line is drawn and touches the semicircle at $B$. An other tangent line at a point, $C$, on the semicircle is drawn. We project the point $C$ on the segment$[AB]$ on a point $D$ . The two tangent lines intersect at the point $T$. + +Prove that the line $(AT)$ bisects $(CD)$ + +\begin{center} +\begin{tkzexample}[] +\begin{tikzpicture}[scale=1.25] + \tkzInit[ymin=-1,ymax=7] + \tkzClip + \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 cm) \tkzGetFirstPoint{C} + \tkzDefLine[orthogonal=through C](I,C) \tkzGetPoint{c} + \tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b} + \tkzInterLL(C,c)(B,b) \tkzGetPoint{T} + \tkzInterLL(A,T)(C,D) \tkzGetPoint{P} + \tkzDrawArc(I,B)(A) + \tkzDrawSegments(A,B A,T C,D I,C) \tkzDrawSegment[color=orange](I,C) + \tkzDrawLine[add = 1 and 0](C,T) \tkzDrawLine[add = 0 and 1](B,T) + \tkzMarkRightAngle(I,C,T) + \tkzDrawPoints(A,B,I,D,C,T) + \tkzLabelPoints(A,B,I,D) \tkzLabelPoints[above right](C,T) + \tkzMarkSegment[pos=.25,mark=s|](C,D) \tkzMarkSegment[pos=.75,mark=s|](C,D) +\end{tikzpicture} +\end{tkzexample} +\end{center} + +\subsection{Example from Dimitris Kapeta} + +You need in this example to use \tkzname{mkpos=.2} with \tkzcname{tkzMarkAngle} because the measure of $ \widehat{CAM}$ is too small. +Another possiblity is to use \tkzcname{tkzFillAngle}. + +\begin{center} +\begin{tkzexample}[] +\begin{tikzpicture}[scale=1.25] + \tkzInit[xmin=-5.2,xmax=3.2,ymin=-3.2,ymax=3.3] + \tkzClip + \tkzDefPoint(0,0){O} + \tkzDefPoint(2.5,0){N} + \tkzDefPoint(-4.2,0.5){M} + \tkzDefPointBy[rotation=center O angle 30](N) + \tkzGetPoint{B} + \tkzDefPointBy[rotation=center O angle -50](N) + \tkzGetPoint{A} + \tkzInterLC(M,B)(O,N) \tkzGetFirstPoint{C} + \tkzInterLC(M,A)(O,N) \tkzGetSecondPoint{A'} + \tkzMarkAngle[fill=blue!25,mkpos=.2, size=0.5](A,C,B) + \tkzMarkAngle[fill=green!25,mkpos=.2, size=0.5](A,M,C) + \tkzDrawSegments(A,C M,A M,B) + \tkzDrawCircle(O,N) + \tkzLabelCircle[above left](O,N)(120){$\mathcal{C}$} + \tkzMarkAngle[fill=red!25,mkpos=.2, size=0.5cm](C,A,M) + \tkzDrawPoints(O, A, B, M, B, C) + \tkzLabelPoints[right](O,A,B) + \tkzLabelPoints[above left](M,C) + \tkzLabelPoint[below left](A'){$A'$} +\end{tikzpicture} +\end{tkzexample} +\end{center} + +\newpage +\subsection{Example 1 from John Kitzmiller } +This figure is the last of beamer document. You can find the document on my site + +Prove $\bigtriangleup LKJ$ is equilateral + +\begin{center} +\begin{tkzexample}[vbox] +\begin{tikzpicture}[scale=1.5] + \tkzDefPoint[label=below left:A](0,0){A} + \tkzDefPoint[label=below right:B](6,0){B} + \tkzDefTriangle[equilateral](A,B) \tkzGetPoint{C} + \tkzMarkSegments[mark=|](A,B A,C B,C) + \tkzDefBarycentricPoint(A=1,B=2) \tkzGetPoint{C'} + \tkzDefBarycentricPoint(A=2,C=1) \tkzGetPoint{B'} + \tkzDefBarycentricPoint(C=2,B=1) \tkzGetPoint{A'} + \tkzInterLL(A,A')(C,C') \tkzGetPoint{J} + \tkzInterLL(C,C')(B,B') \tkzGetPoint{K} + \tkzInterLL(B,B')(A,A') \tkzGetPoint{L} + \tkzLabelPoint[above](C){C} + \tkzDrawPolygon(A,B,C) \tkzDrawSegments(A,J B,L C,K) + \tkzMarkAngles[fill= orange,size=1cm,opacity=.3](J,A,C K,C,B L,B,A) + \tkzLabelPoint[right](J){J} + \tkzLabelPoint[below](K){K} + \tkzLabelPoint[above left](L){L} + \tkzMarkAngles[fill=orange, opacity=.3,thick,size=1,](A,C,J C,B,K B,A,L) + \tkzMarkAngles[fill=green, size=1, opacity=.5](A,C,J C,B,K B,A,L) + \tkzFillPolygon[color=yellow, opacity=.2](J,A,C) + \tkzFillPolygon[color=yellow, opacity=.2](K,B,C) + \tkzFillPolygon[color=yellow, opacity=.2](L,A,B) + \tkzDrawSegments[line width=3pt,color=cyan,opacity=0.4](A,J C,K B,L) + \tkzDrawSegments[line width=3pt,color=red,opacity=0.4](A,L B,K C,J) + \tkzMarkSegments[mark=o](J,K K,L L,J) +\end{tikzpicture} +\end{tkzexample} + +\end{center} + +\newpage +\subsection{Example 2 from John Kitzmiller } +Prove $\dfrac{AC}{CE}=\dfrac{BD}{DF} \qquad$ + +Another interesting example from John, you can see how to use some extra options like \tkzname{decoration} and \tkzname{postaction} from \TIKZ\ with \tkzname{tkz-euclide}. + +\begin{center} +\begin{tkzexample}[vbox] +\begin{tikzpicture}[scale=1.5,decoration={markings, + mark=at position 3cm with {\arrow[scale=2]{>}};}] + \tkzInit[xmin=-0.25,xmax=6.25, ymin=-0.5,ymax=4] + \tkzClip + \tkzDefPoints{0/0/E, 6/0/F, 0/1.8/P, 6/1.8/Q, 0/3/R, 6/3/S} + \tkzDrawLines[postaction={decorate}](E,F P,Q R,S) + \tkzDefPoints{3.5/3/A, 5/3/B} + \tkzDrawSegments(E,A F,B) + \tkzInterLL(E,A)(P,Q) \tkzGetPoint{C} + \tkzInterLL(B,F)(P,Q) \tkzGetPoint{D} + \tkzLabelPoints[above right](A,B) + \tkzLabelPoints[below](E,F) + \tkzLabelPoints[above left](C) + \tkzDrawSegments[style=dashed](A,F) + \tkzInterLL(A,F)(P,Q) \tkzGetPoint{G} + \tkzLabelPoints[above right](D,G) + \tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](A,C A,G) + \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](C,E G,F) + \tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](B,D) + \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](D,F) +\end{tikzpicture} +\end{tkzexample} +\end{center} + +\newpage +\subsection{Example 3 from John Kitzmiller } +Prove $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector) + + +\begin{center} +\begin{tkzexample}[vbox] +\begin{tikzpicture}[scale=1.5] + \tkzInit[xmin=-4,xmax=5,ymax=4.5] \tkzClip[space=.5] + \tkzDefPoints{0/0/B, 5/0/D} \tkzDefPoint(70:3){A} + \tkzDrawPolygon(B,D,A) + \tkzDefLine[bisector](B,A,D) \tkzGetPoint{a} + \tkzInterLL(A,a)(B,D) \tkzGetPoint{C} + \tkzDefLine[parallel=through B](A,C) \tkzGetPoint{b} + \tkzInterLL(A,D)(B,b) \tkzGetPoint{P} + \begin{scope}[decoration={markings, + mark=at position .5 with {\arrow[scale=2]{>}};}] + \tkzDrawSegments[postaction={decorate},dashed](C,A P,B) + \end{scope} + \tkzDrawSegment(A,C) \tkzDrawSegment[style=dashed](A,P) + \tkzLabelPoints[below](B,C,D) \tkzLabelPoints[above](A,P) + \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](B,C P,A) + \tkzDrawSegments[color=teal, line width=3pt, opacity=0.4](C,D A,D) + \tkzDrawSegments[color=magenta, line width=3pt, opacity=0.4](A,B) + \tkzMarkAngles[size=0.7](B,A,C C,A,D) + \tkzMarkAngles[size=0.7, fill=green, opacity=0.5](B,A,C A,B,P) + \tkzMarkAngles[size=0.7, fill=yellow, opacity=0.3](B,P,A C,A,D) + \tkzMarkAngles[size=0.7, fill=green, opacity=0.6](B,A,C A,B,P B,P,A C,A,D) + \tkzLabelAngle[pos=1](B,A,C){1} \tkzLabelAngle[pos=1](C,A,D){2} + \tkzLabelAngle[pos=1](A,B,P){3}) \tkzLabelAngle[pos=1](B,P,A){4} + \tkzMarkSegments[mark=|](A,B A,P) +\end{tikzpicture} +\end{tkzexample} +\end{center} + +\newpage +\subsection{Example 4 from John Kitzmiller } +Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour) + +\begin{center} +\begin{tkzexample}[vbox] +\begin{tikzpicture}[scale=2] + \tkzInit[xmax=5, ymax=5] + \tkzDefPoint(0,3){A} \tkzDefPoint(6,3){E} \tkzDefPoint(1.35,3){B} + \tkzDefPoint(4.65,3){D} \tkzDefPoint(1,1){G} \tkzDefPoint(5,5){F} + \tkzDefMidPoint(A,E) \tkzGetPoint{C} + \tkzFillPolygon[yellow, opacity=0.4](B,G,C) + \tkzFillPolygon[yellow, opacity=0.4](D,F,C) + \tkzFillPolygon[blue, opacity=0.3](A,B,G) + \tkzFillPolygon[blue, opacity=0.3](E,D,F) + \tkzMarkAngles[size=0.6,fill=green](B,G,A D,F,E) + \tkzMarkAngles[size=0.6,fill=orange](B,C,G D,C,F) + \tkzMarkAngles[size=0.6,fill=yellow](G,B,C F,D,C) + \tkzMarkAngles[size=0.6,fill=red](A,B,G E,D,F) + \tkzMarkSegments[mark=|](B,C D,C) \tkzMarkSegments[mark=s||](G,C F,C) + \tkzMarkSegments[mark=o](A,G E,F) \tkzMarkSegments[mark=s](B,G D,F) + \tkzDrawSegment[color=red](A,E) + \tkzDrawSegment[color=blue](F,G) + \tkzDrawSegments(A,G G,B E,F F,D) + \tkzLabelPoints[below](C,D,E,G) \tkzLabelPoints[above](A,B,F) +\end{tikzpicture} +\end{tkzexample} +\end{center} +\endinput
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