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%\usepackage[french,lined,boxed,commentsnumbered]{algorithm2e}
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	Cadre jaune
%\textbf{Nom : }$\dots\dots\dots\dots\dots$ \hfill \textbf{Pr\'enom : }$\dots\dots\dots\dots\dots$  \hfill Classe :$\dots\dots$\\
\psframebox[linestyle=solid,shadow=true,framearc=0.5,fillcolor=vert!30]{
\begin{minipage}{17cm}
\begin{center}
\vspace{1mm}
\textsc{\textbf{Correction de l'interrogation n$^{\circ}14$ }}
\end{center}
\vspace{-3mm}
\end{minipage}
}
\setcounter{section}{1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	 Interro


\Exoi{\hfill (4 points)\\
On considère un triangle $ABC$. \\
On note $A'$ le milieu du segment $[BC]$, $B'$ le milieu du segment $[AC]$ et enfin $C'$ le milieu du segment $[AB]$. \\
On note $G$ le centre de gravité du triangle $ABC$
\begin{enumerate}
\item \quad \\
\begin{center}
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\psdots[dotsize=1pt 0,dotstyle=*](1.85,1.38)
\rput[bl](1.96,1.46){$G$}
\end{pspicture*}
\end{center}
\item Compléter les égalités suivantes :
$$ \overrightarrow{AG}=\dfrac{2}{3} \overrightarrow{AA'}\qquad \text{et}\qquad \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$$
\end{enumerate}}

\Exoi{\hfill (6 points)\\
Soit $ABCD$ un parallélogramme de centre $I$.
\begin{enumerate}
\item \quad \\
\begin{center}
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\rput[bl](-0.42,0.96){$A$}
\psdots[dotsize=1pt 0,dotstyle=*](3.48,0.7)
\rput[bl](3.55,0.73){$B$}
\psdots[dotsize=1pt 0,dotstyle=*](2.31,-0.15)
\rput[bl](2.29,-0.44){$C$}
\psdots[dotsize=1pt 0,dotstyle=*](-1.4,0.01)
\rput[bl](-1.7,0.08){$D$}
\psdots[dotsize=1pt 0,dotstyle=*](7.18,0.54)
\rput[bl](7.25,0.57){$M$}
\psdots[dotsize=1pt 0,dotstyle=*](1.04,0.35)
\rput[bl](1.12,0.38){$I$}
\psdots[dotsize=1pt 0,dotstyle=*](4.84,-1.16)
\rput[bl](4.97,-1.38){$N$}
\psdots[dotsize=1pt 0,dotstyle=*](-2.57,-0.85)
\rput[bl](-2.91,-1.02){$P$}
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\end{center}
\item Comme $M$ est l'image de $B$ par la translation de vecteur $ \overrightarrow{AB}$ on a : $\overrightarrow{BM}=\overrightarrow{AB}$.
\item Placer $N$ tel que $ \overrightarrow{AN}= 4 \overrightarrow{AI}$.
\item Placer $P$ tel que $ \overrightarrow{AP}=2 \overrightarrow{AC}+\overrightarrow{MA}$.
\item D'après la question précédente on a : $$ \overrightarrow{AP}=2 \overrightarrow{AC}+\overrightarrow{MA}\Longleftrightarrow \overrightarrow{MA}+\overrightarrow{AC}=\overrightarrow{AP}-\overrightarrow{AC}\Longleftrightarrow \overrightarrow{MC}=\overrightarrow{AP}+\overrightarrow{CA}=\overrightarrow{CA}+\overrightarrow{AP}=\overrightarrow{CP}$$ 
Par conséquent $C$ est le milieu de $[MP]$.
\item Comme $ABCD$ est un parallélogramme $I$ est le milieu de $[AC]$, par conséquent $ \overrightarrow{AC}=2 \overrightarrow{AI}$.\\Or,
$ \overrightarrow{AN}= 4 \overrightarrow{AI}= 2\times 2\overrightarrow{AI}=2 \overrightarrow{AC}$, ainsi $C$ est le milieu $[AN]$.\\
On en déduit que $AMNP$ a ses diagonales qui se coupent en leurs milieux et donc $AMNP$ est un parallélogramme.
\end{enumerate}}



\newpage


\setcounter{Exi}{0}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	Cadre jaune
%\textbf{Nom : }$\dots\dots\dots\dots\dots$ \hfill \textbf{Pr\'enom : }$\dots\dots\dots\dots\dots$  \hfill Classe :$\dots\dots$\\
\psframebox[linestyle=solid,shadow=true,framearc=0.5,fillcolor=vert!30]{
\begin{minipage}{17cm}
\begin{center}
\vspace{1mm}
\textsc{\textbf{Correction de l'interrogation n$^{\circ}14$ }}
\end{center}
\vspace{-3mm}
\end{minipage}
}

\Exoi{\hfill (4 points)\\
Dire si les affirmations suivantes sont vraies ou fausses :\\
\textbf{Une bonne réponse rapporte $0.5$, une mauvaise réponse enlève $0,25$ et une absence de réponse ne rapporte ni n'enlève de points.}
\begin{multicols}{2}
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\begin{multicols}{2}
\begin{enumerate}
\item $ \overrightarrow{AB}=\overrightarrow{CD}$ \textbf{F}
\item $ \overrightarrow{AI}=\overrightarrow{CI}$ \textbf{F}
\item $ \overrightarrow{DC}+\overrightarrow{DA}=\overrightarrow{DB}$ \textbf{V}
\item $ \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}$ \textbf{V}
\item $ \overrightarrow{CD}=\overrightarrow{BA}$ \textbf{V}
\item $ \overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{0}$ \textbf{V}
\item $ \overrightarrow{AI}=\dfrac{1}{2} \overrightarrow{CA}$ \textbf{F}
\item $ \overrightarrow{DB}=2\overrightarrow{DI}$ \textbf{V}
\end{enumerate}
\end{multicols}
\end{multicols}
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	 Interro bis

\Exoi{\hfill (6 points)\\
Soit $ABC$ un triangle, $A'$ le milieu de $[BC]$.
\begin{enumerate}
\item \quad \\
\begin{center}
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\psdots[dotsize=1pt 0,dotstyle=*](9.49,0.93)
\rput[bl](9.59,0.98){$D$}
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\psdots[dotsize=1pt 0,dotstyle=*](0.54,1.91)
\rput[bl](0.66,1.96){$G$}
\end{pspicture*}
\end{center}
\item Placer $D$ tel que $ \overrightarrow{CD}=2\overrightarrow{AB}$
\item Placer $E$ tel que $ \overrightarrow{BE}=2 \overrightarrow{BA}$
\item On a $ \overrightarrow{BE}=2 \overrightarrow{BA}\Longleftrightarrow \overrightarrow{EB}=2\overrightarrow{AB}=\overrightarrow{CD}$, par conséquent $BDCE$ est un parallélogramme.
\item $A'$ est le milieu de $[BC]$, diagonale du parallélogramme $BDCE$ par conséquent  $A'$ est le centre du parallélogramme $BDCE$.
\item Placer $G$ tel que $ \overrightarrow{AG}=\dfrac{2}{3} \overrightarrow{AA'}$. $G$ est donc le centre de gravité de $ABC$.
\end{enumerate}
}
\end{document}
