\begin{tableau}[pr]{\linewidth}
  \\ \hline %%%%%%%%%%%%%%%%%%%%%%P53
\begin{tikzpicture}[general, yscale=0.2, xscale=0.3]
\coordinate (A) at (2,7);
\coordinate (B) at (0,0);
\coordinate (C) at (10,0);
\coordinate (A') at (5,0);
\coordinate (B') at (6,3.5);
\coordinate (C') at (1,3.5);
\coordinate (G) at (4,2.3);
\draw[color=F1, fill=F4] (A)--(B)--(C)--cycle; 
\draw[dashed] (A)--(A'); 
\draw[dashed] (B)--(B'); 
\draw[dashed] (C)--(C'); 
\foreach \N/\pos in {A/above, B/below left, C/below right, A'/below, B'/above right, C'/left} {\draw (\N) node[\pos] {$\N$};}
\foreach \x/\y/\symb in {0.5/1.75/$\approx$, 1.5/5.25/$\approx$, 4/5.25/$\infty$, 8/1.75/$\infty$, 2.5/0/$\int$, 7.5/0/$\int$} {\draw[color=A1] (\x,\y) node {\small {\boldmath \symb}};}
\end{tikzpicture} &
  \propriete{}
Dans un triangle, le centre de gravité se situe au deux~tiers des médianes en partant du sommet& Ici, $G$ est le centre de gravité du triangle $ABC$ et $[AA']$ en est une médiane. 

Donc $AG=\dfrac{\TopStrut 2}{\BotStrut 3}AA'$.
  \\ \hline
\end{tableau}
