\documentclass[12pt]{article}
\usepackage{ammaths}
\usepackage{pst-eucl}

\begin{document}

\module{Classic configurations}{1}{AP03}{1 period}

\prereq{Configurations and theorems seen during previous years.}

\object{\begin{itemize}
        \item Review the main configurations and theorems of the previous years.
	\item Use these configurations to solve two problems.
	\end{itemize}}

\mater{\begin{itemize}
	\item Answer sheets with three configurations on each.
	\item Problems.
	\end{itemize}}

\modpart{Scrambled groups - part 1}{15 mins}

Students are working in 4 groups of 4 or 5. Each group is given an answer sheet with three classic configurations. They have to find the name and precise description of the situation or concept illustrated.

\modpart{Scrambled groups - part 2}{Remaining time}

Groups are scrambled, so that in each new group there is one member of each of the four original groups. They have to cooperate to solve a problem, where the configurations they've reviewed may (or may not) be useful.

\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Answer sheet 1}


\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Right-angled triangle
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstTriangle[PointSymbol=none](0,0){A}(0,2){B}(3,0){C}
\pstRightAngle{C}{A}{B}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}

\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Two lines perpendicular to a third one.
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{nodesep=-0.3}
\pstGeonode[PointName=none, PointSymbol=none](0.5,0){A}(0.5,1){B}(0.5,2){C}(1.5,0){D}(1.5,1){E}(1.5,2){F}
\pstLineAB{C}{A} \pstLineAB{B}{E} \pstLineAB{D}{F}
\pstRightAngle[RightAngleSize=0.2]{F}{E}{B}
\pstRightAngle[RightAngleSize=0.2]{E}{B}{C}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Medians
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{PointSymbol=none}
\pstTriangle{A}(3,0){B}(2,2.5){C}
\pstMiddleAB[PosAngle=-90, SegmentSymbol=pstslash, PointName=none]{A}{B}{K}
\pstMiddleAB[PosAngle=20, SegmentSymbol=pstslashh]{B}{C}{I}
\pstMiddleAB[PosAngle=150, SegmentSymbol=MarkCros, PointName=none]{C}{A}{J}
\pstLineAB[linestyle=dashed]{A}{I}
\pstLineAB[linestyle=dashed]{B}{J}
\pstLineAB[linestyle=dashed]{C}{K}
\pstCGravABC{A}{B}{C}{G}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}

\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Answer sheet 2}



\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Inscribed angles
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PosAngle={-90,0}](1,1){O}(2.2,1){A}
\pstCircleOA{O}{A}
\pstCurvAbsNode{O}{A}{B}{\pstDistVal{3}}
\pstCurvAbsNode{O}{A}{N}{\pstDistVal{6}}
\pstCurvAbsNode{O}{A}{M}{\pstDistVal{5}}
\pstLineAB{O}{A}
\pstLineAB{O}{B}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{A}{O}{B}{$\theta$}
\psset{linestyle=dashed}
\pstLineAB{M}{A}
\pstLineAB{M}{B}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{A}{M}{B}{$\frac{\theta}{2}$}
\pstLineAB{N}{A}
\pstLineAB{N}{B}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{A}{N}{B}{$\frac{\theta}{2}$}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Angular bisectors
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{PointSymbol=none}
\pstTriangle{A}(3,0){B}(2,2.5){C}
\pstBissectBAC[linestyle=dashed, PointName=none]{B}{A}{C}{A'}
\pstBissectBAC[linestyle=dashed, PointName=none]{A}{C}{B}{C'}
\pstBissectBAC[linestyle=dashed, PointName=none]{C}{B}{A}{B'}
\pstMarkAngle[Mark=MarkHash, MarkAngleRadius=0.8]{B}{A}{A'}{}
\pstMarkAngle[Mark=MarkHash, MarkAngleRadius=0.7]{A'}{A}{C}{}
\pstMarkAngle[MarkAngleRadius=0.8]{C}{B}{B'}{}
\pstMarkAngle[MarkAngleRadius=0.9]{B'}{B}{A}{}
\pstMarkAngle[Mark=MarkCros, MarkAngleRadius=0.7]{A}{C}{C'}{}
\pstMarkAngle[Mark=MarkCros, MarkAngleRadius=0.6]{C'}{C}{B}{}
\pstInterLL{C}{C'}{A}{A'}{I}
\pstProjection[PointName=none]{A}{B}{I}[I']
\pstCircleOA{I}{I'}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Isosceles triangle
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstTriangle[PointSymbol=none](0,0){A}(3,0){B}(1.5,2.5){C}
\psset{SegmentSymbol=MarkCros} \pstMiddleAB[PosAngle=225]{A}{B}{C'}
\pstSegmentMark{A}{C}\pstSegmentMark{C}{B}
\pstLineAB{C}{C'}
\pstRightAngle{B}{C'}{C}
\pstBissectBAC[PointName=none,PointSymbol=none]{A}{C}{B}{C'}
\pstMarkAngle[Mark=MarkHash]{A}{C}{C'}{}
\pstMarkAngle[Mark=MarkHash]{C'}{C}{B}{}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}

\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Answer sheet 3}



\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Perpendicular bisectors
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{PointSymbol=none}
\pstTriangle(0,0.5){A}(2.5,0.5){B}(1.7,2.5){C}
\psset{CodeFig=true}
\pstCircleABC[RightAngleSize=0.2]{A}{B}{C}{O}
\psset{linecolor=black}
\pstTriangle(0,0.5){A}(2.5,0.5){B}(1.7,2.5){C}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Intercept theorem
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PosAngle={-50,-40,80}](1,1){O}(2,1){A}(1.5,2){B}
\pstHomO[HomCoef=-0.5, PosAngle={75,180}]{O}{A,B}[M,N]
\psset{nodesep=-0.5}
\pstLineAB{A}{M}
\pstLineAB{B}{N}
\psset{nodesep=-0.5}
\pstLineAB{A}{B}
\pstLineAB{M}{N}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Thales' theorem
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PosAngle={-90,0,180}](1.5,1.25){O}(2.75,1.25){A}(0.25,1.25){B}
\pstCircleOA{O}{A}
\pstCurvAbsNode{O}{A}{C}{\pstDistVal{2}}
\pstLineAB{A}{B}
\pstLineAB{B}{C}
\pstLineAB{C}{A}
\pstRightAngle{A}{C}{B}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}

\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Answer sheet 4}


\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Alternate-internal angles
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PointName=none, PointSymbol=none](0,0.5){A}(1,0.5){E}(2.5,0.5){B}(0,1.5){C}(2,1.5){F}(2.5,1.5){D}
\psset{nodesep=-0.3}
\pstLineAB{A}{B} \pstLineAB{C}{D} \pstLineAB{E}{F}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{F}{E}{A}{$\theta$}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{E}{F}{D}{$\theta$}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Tangent to a circle
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{nodesep=0}
\pstGeonode[PosAngle={90,0}](1,1){O}(2,1){A}
\pstGeonode[PointName=none, PointSymbol=none](2,2.5){B}(2,0){C}
\pstCircleOA{O}{A}
\pstLineAB{C}{B}
\pstLineAB{O}{A}
\pstRightAngle[RightAngleSize=0.2]{B}{A}{O}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}


\vspace{1cm}

\begin{tabular}{p{6cm}p{9.5cm}}
\parbox{6cm}{
\begin{center}
\psset{unit=1.5cm}
%Altitudes and orthocenter
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{CodeFig=true, PointSymbol=none}
\pstTriangle[PosAngleA=180]{A}(3,0){B}(2,2){C}
\pstProjection[PosAngle=-90, RightAngleSize=0.2]{B}{A}{C}
\pstProjection[RightAngleSize=0.2]{B}{C}{A}
\pstProjection[PosAngle=90,RightAngleSize=0.2]{A}{C}{B}
\pstInterLL[PosAngle=180, PointSymbol=square]{A}{A'}{B}{B'}{H}
\end{pspicture}
\end{center}} &
\parbox{9.5cm}{
{\Large Name :\dotfill\medskip\\ 
Property : \dotfill\medskip\par \dotfill\medskip\par \dotfill\medskip\par
}} \\
\end{tabular}

\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Problems}

Any configuration used in these problems must be explicitely stated. It will also be useful to draw every geometrical object mentioned in the questions, with different colors.


\sectionblack{Problem 1}


Let $ABC$ be a scalene triangle such that $AB$ is greater than $AC$ and $\Co$ the circle of centre $A$ and radius $AC$. The line $AB$ intersects the circle in $D$ and $E$, with $D$ on the segment $AB$. The line $\mathscr{D}$ is parallel to $CE$ and passes through $A$. The aim of this exercise is to prove that $\mathscr{D}$ is the angular bisector of the angle $\angle BAC$.
	\begin{enumerate}
	\item Draw a precise figure.
	\item Use a classic configuration to find out what kind of triangle $DCE$ is.
	\item What can you say about the lines $\mathscr{D}$, $DC$ and $CE$. 
	\item What can you deduce about the line $\mathscr{D}$ in the triangle $ADC$ ?
	\item Find out what kind of triangle $ADC$ is.
	\item Conclude.
	\end{enumerate}

\sectionblack{Problem 2}

Consider a circle $\Co$ of centre $O$ and two points $A$ and $B$ on the circle, such that $AB$ is not a diameter of the circle and $AOB$ is not right-angled. The tangets to the circle through the points $A$ and $B$ meet in $C$. Now, the lines $AC$ and $BO$ meet in $D$ and the lines $BC$ and $AO$ meet in $E$.

Prove that the lines $DE$ and $CO$ are perpendicular.


\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Lesson}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Medians
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{PointSymbol=none}
\pstTriangle{A}(3,0){B}(2,2.5){C}
\pstMiddleAB[PosAngle=-90, SegmentSymbol=pstslash, PointName=none]{A}{B}{K}
\pstMiddleAB[PosAngle=20, SegmentSymbol=pstslashh]{B}{C}{I}
\pstMiddleAB[PosAngle=150, SegmentSymbol=MarkCros, PointName=none]{C}{A}{J}
\pstLineAB[linestyle=dashed]{A}{I}
\pstLineAB[linestyle=dashed]{B}{J}
\pstLineAB[linestyle=dashed]{C}{K}
\pstCGravABC{A}{B}{C}{G}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Medians and the center of gravity\\
Property : The three medians of a triangle meet at the center of gravity.
}} \\
\end{tabular}

\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Altitudes and orthocenter
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{CodeFig=true, PointSymbol=none}
\pstTriangle[PosAngleA=180]{A}(3,0){B}(2,2){C}
\pstProjection[PosAngle=-90, RightAngleSize=0.2]{B}{A}{C}
\pstProjection[RightAngleSize=0.2]{B}{C}{A}
\pstProjection[PosAngle=90,RightAngleSize=0.2]{A}{C}{B}
\pstInterLL[PosAngle=180, PointSymbol=square]{A}{A'}{B}{B'}{H}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Altitudes and orthocenter\\ 
Property : The three altitudes of a triangle meet at the orthocenter of that circle.
}} \\
\end{tabular}

\vspace{0.5cm}


\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Angular bisectors
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{PointSymbol=none}
\pstTriangle{A}(3,0){B}(2,2.5){C}
\pstBissectBAC[linestyle=dashed, PointName=none]{B}{A}{C}{A'}
\pstBissectBAC[linestyle=dashed, PointName=none]{A}{C}{B}{C'}
\pstBissectBAC[linestyle=dashed, PointName=none]{C}{B}{A}{B'}
\pstMarkAngle[Mark=MarkHash, MarkAngleRadius=0.8]{B}{A}{A'}{}
\pstMarkAngle[Mark=MarkHash, MarkAngleRadius=0.7]{A'}{A}{C}{}
\pstMarkAngle[MarkAngleRadius=0.8]{C}{B}{B'}{}
\pstMarkAngle[MarkAngleRadius=0.9]{B'}{B}{A}{}
\pstMarkAngle[Mark=MarkCros, MarkAngleRadius=0.7]{A}{C}{C'}{}
\pstMarkAngle[Mark=MarkCros, MarkAngleRadius=0.6]{C'}{C}{B}{}
\pstInterLL{C}{C'}{A}{A'}{I}
\pstProjection[PointName=none]{A}{B}{I}[I']
\pstCircleOA{I}{I'}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Angular bisectors and the center of the inside circle\\ 
Property : The intersection of the three angular bisectors of a triangle is the center of the inside circle.
}} \\
\end{tabular}


\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Perpendicular bisectors
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{PointSymbol=none}
\pstTriangle(0,0.5){A}(2.5,0.5){B}(1.7,2.5){C}
\psset{CodeFig=true}
\pstCircleABC[RightAngleSize=0.2]{A}{B}{C}{O}
\psset{linecolor=black}
\pstTriangle(0,0.5){A}(2.5,0.5){B}(1.7,2.5){C}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Perpendicular bisectors and the circumcenter\\
Property : The intersection of the three perpendicular bisectors of a triangle is
the center of the circumcircle (also called the circumcenter).
}} \\
\end{tabular}

\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Isosceles triangle
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstTriangle[PointSymbol=none](0,0){A}(3,0){B}(1.5,2.5){C}
\psset{SegmentSymbol=MarkCros} \pstMiddleAB[PosAngle=225]{A}{B}{C'}
\pstSegmentMark{A}{C}\pstSegmentMark{C}{B}
\pstLineAB{C}{C'}
\pstRightAngle{B}{C'}{C}
\pstBissectBAC[PointName=none,PointSymbol=none]{A}{C}{B}{C'}
\pstMarkAngle[Mark=MarkHash]{A}{C}{C'}{}
\pstMarkAngle[Mark=MarkHash]{C'}{C}{B}{}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Isosceles triangle\\ 
Property : In an isosceles triangle, the median, the altitude, the angular bisector and the perpendicular bisector from the main vertex are the same.
}} \\
\end{tabular}

\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Thales' theorem
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PosAngle={-90,0,180}](1.5,1.25){O}(2.75,1.25){A}(0.25,1.25){B}
\pstCircleOA{O}{A}
\pstCurvAbsNode{O}{A}{C}{\pstDistVal{2}}
\pstLineAB{A}{B}
\pstLineAB{B}{C}
\pstLineAB{C}{A}
\pstRightAngle{A}{C}{B}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Thales' theorem\\ 
Property : if $A$, $B$ and $C$ are points on a circle such that $AC$ is a diameter, then the angle $\angle ABC$ is a right angle.
}} \\
\end{tabular}
\pagebreak

\moddocdis{Classic configurations}{1}{AP03}{Lesson}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Alternate-internal angles
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PointName=none, PointSymbol=none](0,0.5){A}(1,0.5){E}(2.5,0.5){B}(0,1.5){C}(2,1.5){F}(2.5,1.5){D}
\psset{nodesep=-0.3}
\pstLineAB{A}{B} \pstLineAB{C}{D} \pstLineAB{E}{F}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{F}{E}{A}{$\theta$}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{E}{F}{D}{$\theta$}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Alternate-internal angles\\
Property : If one line intersects two parallel lines, then the alternate-internal angles are equal.
}} \\
\end{tabular}

\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Two lines perpendicular to a third one.
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{nodesep=-0.3}
\pstGeonode[PointName=none, PointSymbol=none](0.5,0){A}(0.5,1){B}(0.5,2){C}(1.5,0){D}(1.5,1){E}(1.5,2){F}
\pstLineAB{C}{A} \pstLineAB{B}{E} \pstLineAB{D}{F}
\pstRightAngle[RightAngleSize=0.2]{F}{E}{B}
\pstRightAngle[RightAngleSize=0.2]{E}{B}{C}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Two lines perpendicular to a third one\\
Property : If two lines are perpendicular to a third one, then they are parallel to each other.
}} \\
\end{tabular}

\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{2cm}{
\begin{center}
\psset{unit=0.7cm}
%Tangent to a circle
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\psset{nodesep=0}
\pstGeonode[PosAngle={90,0}](1,1){O}(2,1){A}
\pstGeonode[PointName=none, PointSymbol=none](2,2.5){B}(2,0){C}
\pstCircleOA{O}{A}
\pstLineAB{C}{B}
\pstLineAB{O}{A}
\pstRightAngle[RightAngleSize=0.2]{B}{A}{O}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Tangent to a circle\\ 
Property : A line which is perpendicular to a radius and which passes through a point of the circle is a tangent to that circle.\\
}} \\
\end{tabular}

\vspace{0.5cm}

\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Right-angled triangle
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstTriangle[PointSymbol=none](0,0){A}(0,2){B}(3,0){C}
\pstRightAngle{C}{A}{B}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Pythagorean theorem\\
Property : A triangle $ABC$ is right-angled in $A$ if and only if $BC^2=AB^2+AC^2$.
}} \\
\end{tabular}


\vspace{0.5cm}


\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Intercept theorem
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PosAngle={-50,-40,80}](1,1){O}(2,1){A}(1.5,2){B}
\pstHomO[HomCoef=-0.5, PosAngle={75,180}]{O}{A,B}[M,N]
\psset{nodesep=-0.5}
\pstLineAB{A}{M}
\pstLineAB{B}{N}
\psset{nodesep=-0.5}
\pstLineAB{A}{B}
\pstLineAB{M}{N}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Intercept theorem\\
Property : Let $BN$ and $AM$ be two lines that meet in $O$. If the lines $AB$ and $MN$ are parallel and if the points $B$, $O$, $N$ are in the same order than $A$, $O$, $M$ then
$$\dfrac{OA}{OM}=\dfrac{OB}{ON}=\dfrac{AB}{MN}$$
}} \\
\end{tabular}

\vspace{0.5cm}


\begin{tabular}{p{3cm}p{12cm}}
\parbox{3cm}{
\begin{center}
\psset{unit=0.7cm}
%Inscribed angles
\begin{pspicture}(-0.5,-0.5)(3.5,2.7)
\pstGeonode[PosAngle={-90,0}](1,1){O}(2.2,1){A}
\pstCircleOA{O}{A}
\pstCurvAbsNode{O}{A}{B}{\pstDistVal{3}}
\pstCurvAbsNode{O}{A}{N}{\pstDistVal{6}}
\pstCurvAbsNode{O}{A}{M}{\pstDistVal{5}}
\pstLineAB{O}{A}
\pstLineAB{O}{B}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{A}{O}{B}{$\theta$}
\psset{linestyle=dashed}
\pstLineAB{M}{A}
\pstLineAB{M}{B}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{A}{M}{B}{$\frac{\theta}{2}$}
\pstLineAB{N}{A}
\pstLineAB{N}{B}
\pstMarkAngle[MarkAngleRadius=0.2, LabelSep=0.4]{A}{N}{B}{$\frac{\theta}{2}$}
\end{pspicture}
\end{center}} &
\parbox{12cm}{
{Name : Inscribed angle\\ 
Property : In a circle, an inscribed angle measures half of the central angle.
}} \\
\end{tabular}



\end{document}
