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\begin{document}
\noindent
\begin{center}
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            \multirow{2}{*}{\parbox{\linewidth}{Prof MOSAID \newline \mylink }}
            & \Centering {Correction Devoir à Domicile 1/Exercice 2} & \hfill  TCS \\
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\noindent
\textbf{\underline{Exercice 2:}}\\
1. La figure:\\
$\overrightarrow{BD}=\frac{2}{3}\overrightarrow{BC}$\\
$\overrightarrow{AE}=-2\overrightarrow{AD}$\\
$\overrightarrow{BF}=\frac{3}{5}\overrightarrow{BE}$\\[7cm]
\hspace*{4cm}
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\stamp{9}{6}\\
2. Montrer que $\overrightarrow{EA}=2\overrightarrow{AB}+\dfrac{4}{3}\overrightarrow{BC}$ puis
$\overrightarrow{FB}=\dfrac{9}{5}\overrightarrow{AB}+\dfrac{4}{5}\overrightarrow{BC}$\\

\begin{minipage}{0.3\textwidth}
\noindent On a:\\
$\overrightarrow{AE}=-2\overrightarrow{AD}$\\
$\overrightarrow{EA}=2\overrightarrow{AD}$\\
$\overrightarrow{EA}=2(\overrightarrow{A\textcolor{red}{B}}+\overrightarrow{\textcolor{red}{B}D})$\\
$\overrightarrow{EA}=2\overrightarrow{AB}+2\overrightarrow{BD}$\\
$\overrightarrow{EA}=2\overrightarrow{AB}+2 \times \textcolor{red}{\frac{2}{3}\overrightarrow{BC}}$\\
$\overrightarrow{EA}=2\overrightarrow{AB}+\frac{4}{3}\overrightarrow{BC}$\\
\end{minipage}
\hspace*{0.2cm}\vline\hspace*{0.2cm}
\begin{minipage}{0.3\textwidth}
\noindent On a:\\
$\overrightarrow{BF}=\frac{3}{5}\overrightarrow{BE}$\\
$\overrightarrow{FB}=\frac{3}{5}\overrightarrow{EB}$\\
$\overrightarrow{FB}=\frac{3}{5}(\overrightarrow{E\textcolor{red}{A}}+\overrightarrow{\textcolor{red}{A}B})$\\
$\overrightarrow{FB}=\frac{3}{5}\overrightarrow{EA}+\frac{3}{5}\overrightarrow{AB}$\\
$\overrightarrow{FB}=\frac{3}{5}(\textcolor{red}{2\overrightarrow{AB}+\frac{4}{3}\overrightarrow{BC}})+\frac{3}{5}\overrightarrow{AB}$\\
$\overrightarrow{FB}=\frac{6}{5}\overrightarrow{AB}+\frac{3}{5}\times \frac{4}{3}\overrightarrow{BC}+\frac{3}{5}\overrightarrow{AB}$\\
$\overrightarrow{FB}=\frac{9}{5}\overrightarrow{AB}+\frac{4}{5}\overrightarrow{BC}$\\
\end{minipage}
\hspace*{0.2cm}\vline\hspace*{0.2cm}
\begin{minipage}{0.3\textwidth}
    Pour montrer que les points A, F et C sont alignés, il suffit de montrer que
    les vecteurs $\overrightarrow{AC}$ et $\overrightarrow{AF}$ sont colinèaires\\
    On va utiliser les deux relations qu'on vient de démontrer:\\
    \stamp{3}{-2}
\end{minipage}
\noindent On a:~
$\overrightarrow{FB}=\frac{9}{5}\overrightarrow{AB}+\frac{4}{5}\overrightarrow{BC}$~~donc~~
$\overrightarrow{FA}+\overrightarrow{AB}=\frac{9}{5}\overrightarrow{AB}+\frac{4}{5}(\overrightarrow{BA}
+\overrightarrow{AC})$\\
Donc~~
$\overrightarrow{FA}=-\overrightarrow{AB}+\frac{9}{5}\overrightarrow{AB}-\frac{4}{5}\overrightarrow{AB}
+\frac{4}{5}\overrightarrow{AC}$\\
Donc~~
$\overrightarrow{FA}=-\frac{5}{5}\overrightarrow{AB}+\frac{9}{5}\overrightarrow{AB}-\frac{4}{5}\overrightarrow{AB}
+\frac{4}{5}\overrightarrow{AC}$
~~donc~~
$\textcolor{red}{\mathbf{\overrightarrow{FA}=\frac{4}{5}\overrightarrow{AC}}}$\\
Alors les vecteurs $\overrightarrow{AC}$ et $\overrightarrow{FA}$ sont colinèaires
et les points A, F et C sont alignés. \\
\textcolor{white}{.}\hfill \underline{MOSAID le \today}\\
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