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+\section{Des exemples}
+\subsection{Quelques exemples intéressants}
+
+\subsubsection{Triangles isocèles semblables}
+
+Ce qui suit provient de l'excellent site \textbf{Descartes et les Mathématiques}. Je n'ai pas modifié le texte et je ne suis l'auteur que de la programmation des figures.
+
+\url{http://debart.pagesperso-orange.fr/seconde/triangle.html}
+
+Bibliographie : Géométrie au Bac - Tangente, hors série no 8 - Exercice 11, page 11
+
+Élisabeth Busser et Gilles Cohen : 200 nouveaux problèmes du Monde - POLE 2007
+
+Affaire de logique n° 364 - Le Monde 17 février 2004
+
+
+Deux énoncés ont été proposés, l'un par la revue \emph{Tangente}, et l'autre par le journal \emph{Le Monde}.
+
+\vspace*{2cm}
+\emph{Rédaction de la revue Tangente} : \textcolor{orange}{On construit deux triangles isocèles semblables AXB et BYC de sommets principaux X et Y, tels que A, B et C soient alignés et que ces triangles soient « indirect ». Soit $\alpha$ l'angle au sommet $\widehat{AXB}$ = $\widehat{BYC}$. On construit ensuite un troisième triangle isocèle XZY semblable aux deux premiers, de sommet principal Z et « indirect ».\\
+On demande de démontrer que le point Z appartient à la droite (AC).}
+
+\vspace*{2cm}
+\emph{Rédaction du Monde} : \textcolor{orange}{On construit deux triangles isocèles semblables AXB et BYC de sommets principaux X et Y, tels que A, B et C soient alignés et que ces triangles soient « indirect ». Soit $\alpha$ l'angle au sommet $\widehat{AXB}$ = $\widehat{BYC}$. Le point Z du segment [AC] est équidistant des deux sommets X et Y.\\
+Sous quel angle voit-il ces deux sommets ?}
+
+\vspace*{2cm} Les constructions et leurs codes associés sont sur les deux pages suivantes, mais vous pouvez chercher avant de regarder. La programmation respecte (il me semble ...), mon raisonnement dans les deux cas.
+\newpage
+
+ \subsubsection{version revue "Tangente"}
+\begin{tkzexample}[]
+\begin{tikzpicture}[scale=.8,rotate=60]
+ \tkzDefPoint(6,0){X} \tkzDefPoint(3,3){Y}
+ \tkzDefShiftPoint[X](-110:6){A} \tkzDefShiftPoint[X](-70:6){B}
+ \tkzDefShiftPoint[Y](-110:4.2){A'} \tkzDefShiftPoint[Y](-70:4.2){B'}
+ \tkzDefPointBy[translation= from A' to B ](Y) \tkzGetPoint{Y}
+ \tkzDefPointBy[translation= from A' to B ](B') \tkzGetPoint{C}
+ \tkzInterLL(A,B)(X,Y) \tkzGetPoint{O}
+ \tkzDefMidPoint(X,Y) \tkzGetPoint{I}
+ \tkzDefPointWith[orthogonal](I,Y)
+ \tkzInterLL(I,tkzPointResult)(A,B) \tkzGetPoint{Z}
+ \tkzDefCircle[circum](X,Y,B) \tkzGetPoint{O}
+ \tkzDrawCircle(O,X)
+ \tkzDrawLines[add = 0 and 1.5](A,C) \tkzDrawLines[add = 0 and 3](X,Y)
+ \tkzDrawSegments(A,X B,X B,Y C,Y) \tkzDrawSegments[color=red](X,Z Y,Z)
+ \tkzDrawPoints(A,B,C,X,Y,O,Z)
+ \tkzLabelPoints(A,B,C,Z) \tkzLabelPoints[above right](X,Y,O)
+\end{tikzpicture}
+\end{tkzexample}
+\newpage
+\subsubsection{version "Le Monde"}
+
+\begin{tkzexample}[]
+\begin{tikzpicture}[scale=1.25]
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(3,0){B}
+ \tkzDefPoint(9,0){C}
+ \tkzDefPoint(1.5,2){X}
+ \tkzDefPoint(6,4){Y}
+ \tkzDefCircle[circum](X,Y,B) \tkzGetPoint{O}
+ \tkzDefMidPoint(X,Y) \tkzGetPoint{I}
+ \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}
+ \tkzDefPointWith[orthogonal](B,Z) \tkzGetPoint{b}
+ \tkzDrawCircle(O,B)
+ \tkzDrawLines[add = 0 and 2,color=orange](B,b)
+ \tkzDrawSegments(A,X B,X B,Y C,Y A,C X,Y)
+ \tkzDrawSegments[color=red](X,Z Y,Z)
+ \tkzDrawPoints(A,B,C,X,Y,Z,M,I)
+ \tkzLabelPoints(A,B,C,Z)
+ \tkzLabelPoints[above right](X,Y,M,I)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Hauteurs d'un triangle}
+
+Ce qui suit provient encore de l'excellent site \textbf{Descartes et les Mathématiques}.
+
+\url{http://debart.pagesperso-orange.fr/geoplan/geometrie_triangle.html}
+
+Les trois hauteurs d'un triangle sont concourantes au même point H.
+
+\begin{tkzexample}[latex=7cm]
+\begin{tikzpicture}[scale=.8]
+ \tkzDefPoint(0,0){C}
+ \tkzDefPoint(7,0){B}
+ \tkzDefPoint(5,6){A}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDefMidPoint(C,B)
+ \tkzGetPoint{I}
+ \tkzDrawArc(I,B)(C)
+ \tkzInterLC(A,C)(I,B)
+ \tkzGetSecondPoint{B'}
+ \tkzInterLC(A,B)(I,B)
+ \tkzGetFirstPoint{C'}
+ \tkzInterLL(B,B')(C,C')
+ \tkzGetPoint{H}
+ \tkzInterLL(A,H)(C,B)
+ \tkzGetPoint{A'}
+ \tkzDefCircle[circum](A,B',C')
+ \tkzGetPoint{O}
+ \tkzDrawCircle[color=red](O,A)
+ \tkzDrawSegments[color=orange](B,B' C,C' A,A')
+ \tkzMarkRightAngles(C,B',B B,C',C C,A',A)
+ \tkzDrawPoints(A,B,C,A',B',C',H)
+ \tkzLabelPoints(A,B,C,A',B',C',H)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Hauteurs - autre construction}
+
+\begin{tkzexample}[latex=7cm]
+\begin{tikzpicture}[scale=.75]
+ \tkzDefPoint(0,0){A}
+ \tkzDefPoint(8,0){B}
+ \tkzDefPoint(3.5,10){C}
+ \tkzDefMidPoint(A,B)
+ \tkzGetPoint{O}
+ \tkzDefPointBy[projection=onto A--B](C)
+ \tkzGetPoint{P}
+ \tkzInterLC(C,A)(O,A)
+ \tkzGetSecondPoint{M}
+ \tkzInterLC(C,B)(O,A)
+ \tkzGetFirstPoint{N}
+ \tkzInterLL(B,M)(A,N)
+ \tkzGetPoint{I}
+ \tkzDrawCircle[diameter](A,B)
+ \tkzDrawSegments(C,A C,B A,B B,M A,N)
+ \tkzMarkRightAngles[fill=brown!20](A,M,B A,N,B A,P,C)
+ \tkzDrawSegment[style=dashed,color=orange](C,P)
+ \tkzLabelPoints(O,A,B,P)
+ \tkzLabelPoint[left](M){$M$}
+ \tkzLabelPoint[right](N){$N$}
+ \tkzLabelPoint[above](C){$C$}
+ \tkzLabelPoint[above right](I){$I$}
+ \tkzDrawPoints[color=red](M,N,P,I)
+ \tkzDrawPoints[color=brown](O,A,B,C)
+\end{tikzpicture}
+\end{tkzexample}
+
+\newpage
+\subsection{Different authors}
+
+\subsubsection{ Square root of the integers }
+How to get $1$, $\sqrt{2}$, $\sqrt{3}$ with a rule and a compass.
+
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}[scale=1.5]
+ \tkzDefPoint(0,0){O}
+ \tkzDefPoint(1,0){a0}
+ \tkzDrawSegment[blue](O,a0)
+ \foreach \i [count=\j] in {0,...,10}{%
+ \tkzDefPointWith[orthogonal normed](a\i,O)
+ \tkzGetPoint{a\j}
+ \tkzDrawPolySeg[color=blue](a\i,a\j,O)}
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+\subsubsection{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{tkzexample}[]
+\begin{tikzpicture}
+ \edef\alphaR{\fpeval{asin(2/3)}}
+ \edef\xB{8-3*cos(\alphaR)}
+ \tkzDrawX[noticks,label=$(d)$]
+ \tkzDefPoint["$A$" above right](8,2){A}
+ \tkzDefPoint[color=red,"$O$" above right](0,0){O}
+ \tkzDefPoint["$B$" above left](\xB,4){B}
+ \tkzDefLine[orthogonal=through B](A,B) \tkzGetPoint{b}
+ \tkzDefPoint(1,0){i}
+ \tkzInterLL(B,b)(O,i) \tkzGetPoint{B'}
+ \tkzDrawSegment[line width=1pt](A,B)
+ \tkzHLine[color=red,style=dashed]{4}
+ \tkzText[above](12,4){$\delta$}
+ \tkzDrawCircle[R,color=blue,line width=.8pt](A,3 cm)
+ \tkzDrawPoint(B')
+ \tkzDrawLine(B,B')
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+\subsubsection{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{tkzexample}[latex=7cm]
+\begin{tikzpicture}
+ \tkzDefPoint["$A$" left](2,1){A}
+ \tkzDefPoint(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}
+
+
+\subsubsection{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{tkzexample}[]
+\begin{tikzpicture}[scale=1.25]
+ \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}
+
+\subsubsection{Exemple : 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{tkzexample}[]
+\begin{tikzpicture}[scale=1.25]
+ \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[mkpos=.2, size=0.5](A,C,B)
+ \tkzMarkAngle[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[mkpos=.2, size=1.2](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}
+
+
+\subsubsection{Example : John Kitzmiller }
+
+Prove $\bigtriangleup LKJ$ is equilateral
+
+
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}[scale=2]
+ \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}
+
+\subsubsection{Exemple : 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{tkzexample}[vbox,small]
+\begin{tikzpicture}[scale=2,decoration={markings,
+ mark=at position 3cm with {\arrow[scale=2]{>}}}]
+ \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}
+
+\subsubsection{Exemple : John Kitzmiller }
+Prove $\dfrac{BC}{CD}=\dfrac{AB}{AD} \qquad$ (Angle Bisector)
+
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}[scale=2]
+ \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}
+
+
+\subsubsection{Exemple : author John Kitzmiller }
+Prove $\overline{AG}\cong\overline{EF} \qquad$ (Detour)
+
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}[scale=2]
+ \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}
+
+\subsubsection{Example from Indonesia}
+
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}[scale=3]
+ \tkzDefPoints{0/0/A,2/0/B}
+ \tkzDefSquare(A,B) \tkzGetPoints{C}{D}
+ \tkzDefPointBy[rotation=center D angle 45](C)\tkzGetPoint{G}
+ \tkzDefSquare(G,D)\tkzGetPoints{E}{F}
+ \tkzInterLL(B,C)(E,F)\tkzGetPoint{H}
+ \tkzFillPolygon[gray!10](D,E,H,C,D)
+ \tkzDrawPolygon(A,...,D)\tkzDrawPolygon(D,...,G)
+ \tkzDrawSegment(B,E)
+ \tkzMarkSegments[mark=|,size=3pt,color=gray](A,B B,C C,D D,A E,F F,G G,D D,E)
+ \tkzMarkSegments[mark=||,size=3pt,color=gray](B,E E,H)
+ \tkzLabelPoints[left](A,D)
+ \tkzLabelPoints[right](B,C,F,H)
+ \tkzLabelPoints[above](G)\tkzLabelPoints[below](E)
+ \tkzMarkRightAngles(D,A,B D,G,F)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{Another example from Indonesia}
+\begin{tkzexample}[vbox,small]
+ \begin{tikzpicture}[pol/.style={fill=brown!40,opacity=.5},
+ seg/.style={tkzdotted,color=gray},
+ hidden pt/.style={fill=gray!40},
+ mra/.style={color=gray!70,tkzdotted,/tkzrightangle/size=.2},
+ scale=3]
+ \tkzSetUpPoint[size=2]
+ \tkzDefPoints{0/0/A,2.5/0/B,1.33/0.75/D,0/2.5/E,2.5/2.5/F}
+ \tkzDefLine[parallel=through D](A,B) \tkzGetPoint{I1}
+ \tkzDefLine[parallel=through B](A,D) \tkzGetPoint{I2}
+ \tkzInterLL(D,I1)(B,I2) \tkzGetPoint{C}
+ \tkzDefLine[parallel=through E](A,D) \tkzGetPoint{I3}
+ \tkzDefLine[parallel=through D](A,E) \tkzGetPoint{I4}
+ \tkzInterLL(E,I3)(D,I4) \tkzGetPoint{H}
+ \tkzDefLine[parallel=through F](E,H) \tkzGetPoint{I5}
+ \tkzDefLine[parallel=through H](E,F) \tkzGetPoint{I6}
+ \tkzInterLL(F,I5)(H,I6) \tkzGetPoint{G}
+ \tkzDefMidPoint(G,H) \tkzGetPoint{P}
+ \tkzDefMidPoint(G,C) \tkzGetPoint{Q}
+ \tkzDefMidPoint(B,C) \tkzGetPoint{R}
+ \tkzDefMidPoint(A,B) \tkzGetPoint{S}
+ \tkzDefMidPoint(A,E) \tkzGetPoint{T}
+ \tkzDefMidPoint(E,H) \tkzGetPoint{U}
+ \tkzDefMidPoint(A,D) \tkzGetPoint{M}
+ \tkzDefMidPoint(D,C) \tkzGetPoint{N}
+ \tkzInterLL(B,D)(S,R) \tkzGetPoint{L}
+ \tkzInterLL(H,F)(U,P) \tkzGetPoint{K}
+ \tkzDefLine[parallel=through K](D,H) \tkzGetPoint{I7}
+ \tkzInterLL(K,I7)(B,D) \tkzGetPoint{O}
+
+ \tkzFillPolygon[pol](P,Q,R,S,T,U)
+ \tkzDrawSegments[seg](K,O K,L P,Q R,S T,U
+ C,D H,D A,D M,N B,D)
+ \tkzDrawSegments(E,H B,C G,F G,H G,C Q,R S,T U,P H,F)
+ \tkzDrawPolygon(A,B,F,E)
+ \tkzDrawPoints(A,B,C,E,F,G,H,P,Q,R,S,T,U,K)
+ \tkzDrawPoints[hidden pt](M,N,O,D)
+ \tkzMarkRightAngle[mra](L,O,K)
+ \tkzMarkSegments[mark=|,size=1pt,thick,color=gray](A,S B,S B,R C,R
+ Q,C Q,G G,P H,P
+ E,U H,U E,T A,T)
+
+ \tkzLabelAngle[pos=.3](K,L,O){$\alpha$}
+ \tkzLabelPoints[below](O,A,S,B)
+ \tkzLabelPoints[above](H,P,G)
+ \tkzLabelPoints[left](T,E)
+ \tkzLabelPoints[right](C,Q)
+ \tkzLabelPoints[above left](U,D,M)
+ \tkzLabelPoints[above right](L,N)
+ \tkzLabelPoints[below right](F,R)
+ \tkzLabelPoints[below left](K)
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+\subsubsection{Three circles}
+
+\begin{tkzexample}[vbox,small]
+\begin{tikzpicture}[scale=1.5]
+ \tkzDefPoints{0/0/A,8/0/B,0/4/a,8/4/b,8/8/c}
+ \tkzDefTriangle[equilateral](A,B) \tkzGetPoint{C}
+ \tkzDrawPolygon(A,B,C)
+ \tkzDefSquare(A,B) \tkzGetPoints{D}{E}
+ \tkzClipBB
+ \tkzDefMidPoint(A,B) \tkzGetPoint{M}
+ \tkzDefMidPoint(B,C) \tkzGetPoint{N}
+ \tkzDefMidPoint(A,C) \tkzGetPoint{P}
+ \tkzDrawSemiCircle[gray,dashed](M,B)
+ \tkzDrawSemiCircle[gray,dashed](A,M)
+ \tkzDrawSemiCircle[gray,dashed](A,B)
+ \tkzDrawCircle[gray,dashed](B,A)
+ \tkzInterLL(A,N)(M,a) \tkzGetPoint{Ia}
+ \tkzDefPointBy[projection = onto A--B](Ia)
+ \tkzGetPoint{ha}
+ \tkzDrawCircle[gray](Ia,ha)
+ \tkzInterLL(B,P)(M,b) \tkzGetPoint{Ib}
+ \tkzDefPointBy[projection = onto A--B](Ib)
+ \tkzGetPoint{hb}
+ \tkzDrawCircle[gray](Ib,hb)
+ \tkzInterLL(A,c)(M,C) \tkzGetPoint{Ic}
+ \tkzDefPointBy[projection = onto A--C](Ic)
+ \tkzGetPoint{hc}
+ \tkzDrawCircle[gray](Ic,hc)
+ \tkzInterLL(A,Ia)(B,Ib) \tkzGetPoint{G}
+ \tkzDrawCircle[gray,dashed](G,Ia)
+ \tkzDrawPolySeg(A,E,D,B)
+ \tkzDrawPoints(A,B,C)
+ \tkzDrawPoints(G,Ia,Ib,Ic)
+ \tkzDrawSegments[gray,dashed](C,M A,N B,P M,a M,b A,a a,b b,B A,D Ia,ha)
+\end{tikzpicture}
+\end{tkzexample}
+
+\subsubsection{"The" Circle of APOLLONIUS}
+
+\begin{tkzexample}[vbox,small]
+ \begin{tikzpicture}[scale=.5]
+ \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[spieker](A,B,C) \tkzGetPoint{Sp}
+ \tkzDefExCircle(A,B,C) \tkzGetPoint{Jb}
+ \tkzDefExCircle(C,A,B) \tkzGetPoint{Ja}
+ \tkzDefExCircle(B,C,A) \tkzGetPoint{Jc}
+ \tkzDefPointBy[projection=onto B--C ](Jc) \tkzGetPoint{Xc}
+ \tkzDefPointBy[projection=onto B--C ](Jb) \tkzGetPoint{Xb}
+ \tkzDefPointBy[projection=onto A--B ](Ja) \tkzGetPoint{Za}
+ \tkzDefPointBy[projection=onto A--B ](Jb) \tkzGetPoint{Zb}
+ \tkzDefLine[parallel=through Xc](A,C) \tkzGetPoint{X'c}
+ \tkzDefLine[parallel=through Xb](A,B) \tkzGetPoint{X'b}
+ \tkzDefLine[parallel=through Za](C,A) \tkzGetPoint{Z'a}
+ \tkzDefLine[parallel=through Zb](C,B) \tkzGetPoint{Z'b}
+ \tkzInterLL(Xc,X'c)(A,B) \tkzGetPoint{B'}
+ \tkzInterLL(Xb,X'b)(A,C) \tkzGetPoint{C'}
+ \tkzInterLL(Za,Z'a)(C,B) \tkzGetPoint{A''}
+ \tkzInterLL(Zb,Z'b)(C,A) \tkzGetPoint{B''}
+ \tkzDefPointBy[reflection= over Jc--Jb](B') \tkzGetPoint{Ca}
+ \tkzDefPointBy[reflection= over Jc--Jb](C') \tkzGetPoint{Ba}
+ \tkzDefPointBy[reflection= over Ja--Jb](A'')\tkzGetPoint{Bc}
+ \tkzDefPointBy[reflection= over Ja--Jb](B'')\tkzGetPoint{Ac}
+ \tkzDefCircle[circum](Ac,Ca,Ba) \tkzGetPoint{Q}
+ \tkzDrawCircle[circum](Ac,Ca,Ba)
+ \tkzDefPointWith[linear,K=1.1](Q,Ac) \tkzGetPoint{nAc}
+ \tkzClipCircle[through](Q,nAc)
+ \tkzDrawLines[add=1.5 and 1.5,dashed](A,B B,C A,C)
+ \tkzDrawPolygon[color=blue](A,B,C)
+ \tkzDrawPolygon[dashed,color=blue](Ja,Jb,Jc)
+ \tkzDrawCircles[ex](A,B,C B,C,A C,A,B)
+ \tkzDrawLines[add=0 and 0,dashed](Ca,Bc B,Za A,Ba B',C')
+ \tkzDrawLine[add=1 and 1,dashed](Xb,Xc)
+ \tkzDrawLine[add=7 and 3,blue](O,K)
+ \tkzDrawLine[add=8 and 15,red](N,Sp)
+ \tkzDrawLines[add=10 and 10](K,O N,Sp)
+ \tkzDrawSegments(Ba,Ca Bc,Ac)
+ \tkzDrawPoints(A,B,C,N,Ja,Jb,Jc,Xb,Xc,B',C',Za,Zb,Ba,Ca,Bc,Ac,Q,Sp,K,O)
+ \tkzLabelPoints(A,B,C,N,Ja,Jb,Jc,Xb,Xc,B',C',Za,Zb,Ba,Ca,Bc,Ac,Q,Sp)
+ \tkzLabelPoints[above](K,O)
+ \end{tikzpicture}
+\end{tkzexample}
+
+
+
+\endinput