\begin{propriete}[Propriété du parallélogramme]
\parbox{0.49\textwidth}{
 Soit $A$, $B$, $C$, $D$ quatre points. 

Dire que $\vv{AD}=\vv{AB}+\vv{AC}$ \'equivaut \`a dire que $ABDC$ est un parall\'elogramme. 
}\hfill
\parbox{0.35\textwidth}{
\begin{tikzpicture}[general, scale=0.3]
\draw [quadrillage] (-1,-3) grid (9,5); 
\draw (1,2) node[left] {{\boldmath $\vv{AB}$}};
 \draw (5,3) node[above] {{\boldmath $\vv{AC}$}};
 \draw (4,2) node[below] {\rotatebox{-56.0}{{\boldmath $\vv{AB}+\vv{AC}$}}};
 \draw (0,0) node[below] {$B$};
 \draw (2,4) node[above] {$A$};
 \draw (8,2) node[above] {$C$};
 \draw (6,-2) node[right] {$D$};
\draw[<-, color=G1, epais] (0,0)--(2,4);
\draw[->, color=A1, epais] (2,4)--(6,-2);
\draw[->, color=C1, epais] (2,4)--(8,2);
\end{tikzpicture} 
}
\end{propriete}

\begin{preuve}
 
On suppose que $\vv{AD}=\vv{AB}+\vv{AC}$.

On utilise la relation de Chasles pour d\'ecomposer $\vv{AD}$: 
$\vv{AD}=\vv{AC}+\vv{CD}=\vv{AB}+\vv{AC}$.

On ajoute $\vv{CA}$ aux deux membres de l'\'egalit\'e :
$\underbrace{\vv{CA}+\vv{AC}}+\vv{CD}=\vv{AB}+\underbrace{\vv{AC}+\vv{CA}}$.

On utilise \`a nouveau la relation de Chasles avec $\vv{CA}+\vv{AC}=\vv 0$ et $\vv{AC}+\vv{CA}=\vv 0$.

Donc, la relation de d\'epart est \'equivalente \`a:
$\vv{CD}=\vv{AB}$ et $ABDC$ est un parall\'elogramme.

\end{preuve}
