%@Auteur:Arnaud Gazagnes\par On considère la figure ci-dessous dans laquelle le triangle $ABC$ est quelconque. \begin{center} {\psset{unit=0.75cm} \begin{pspicture*}(0,0)(15,8) \pstGeonode[PointSymbol=none,PosAngle={180,90,45}](1,1){A}(12,0){B}(3,7){C} \pstLineAB{A}{B}\pstLineAB{A}{C}\pstLineAB{B}{C} \pstMiddleAB[PointName=none,PosAngle=140]{A}{B}{C'}\pstMiddleAB[PointName=none]{B}{C}{A'} \pstSegmentMark[SegmentSymbol=MarkHash]{A}{C'} \pstSegmentMark[SegmentSymbol=MarkHash]{B}{C'} \pstSegmentMark[SegmentSymbol=MarkHashh]{B}{A'} \pstSegmentMark[]{C}{A'} {\psset{CodeFig=true,CodeFigColor=black,nodesepA=-15,nodesepB=-5,PointName=none,PointSymbolB=none,SegmentSymbol=none}\pstMediatorAB{C}{B}{A'}{M_A} \pstMediatorAB{A}{B}{C'}{M_C}} \pstProjection[PointName=none]{A}{C}{B}[H_B] \pstLineAB[nodesepA=-10,nodesepB=-5]{B}{H_B} \pstRightAngle{B}{H_B}{A} \pstLineAB[nodesepA=-5,nodesepB=-5]{C}{C'} \pstProjection[PointName=none]{B}{C}{A}[H_A] \pstLineAB[nodesepA=-5,nodesepB=-5]{A}{H_A} \pstRightAngle{A}{H_A}{C} \pstInterLL[PosAngle=0]{C'}{M_C}{A'}{M_A}{X} \pstInterLL[PosAngle=280]{B}{H_B}{A'}{M_A}{Y} \pstInterLL[PosAngle=-45]{C'}{M_C}{B}{H_B}{Z} \pstInterLL[PosAngle=225]{C'}{C}{B}{H_B}{T} \rput(2,7.5){\psovalbox{\texttt{{\large 1}}}} \rput(5,7){\psovalbox{\texttt{{\large 2}}}} \rput(7.7,7){\psovalbox{\texttt{{\large 3}}}} \rput(10.2,6){\psovalbox{\texttt{{\large 4}}}} \end{pspicture*}} \end{center} \begin{enumerate} \item Explicite chacune des droites suivantes pour le triangle $ABC$. \begin{center} \begin{tabular}{p{.475\textwidth}p{.475\textwidth}} \psovalbox{\texttt{{\large 1}}}&\psovalbox{\texttt{{\large 2}}}\\ \psovalbox{\texttt{{\large 3}}}&\psovalbox{\texttt{{\large 4}}}\\ \end{tabular} \end{center} \item Quel est le centre du cercle circonscrit au triangle $ABC$ ? \emph{(Justifie.)} \item Que peux-tu dire des droites \psovalbox{\texttt{{\large 2}}} et \psovalbox{\texttt{{\large 4}}} ? \emph{(Justifie.)} \end{enumerate}