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-%!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 \ No newline at end of file