
\begin{enumerate}
\item
\begin{itemize}
\item $\begin{aligned}[t]
\vv{AE} + 2\vv{BE}-\vv{CE}=\vv{0} & \iff \vv{AE} + 2\big(\vv{BA}+\vv{AE}\big)-\big(\vv{CA}+\vv{AE}\big)=\vv{0}\\
& \iff 2\vv{AE} = 2\vv{AB} + \vv{CA}\\
& \iff \boxed{\vv{AE} = \vv{AB} + \frac{1}{2}\vv{CA}}
\end{aligned}$

\item $\begin{aligned}[t]
\vv{BF}=\vv{AF}+\vv{CF} & \iff \vv{BF} = \big(\vv{AB}+\vv{BF}\big) + \big(\vv{CB}+\vv{BF}\big)\\
& \iff \vv{0} = \vv{AB}+\vv{BF}  +  \vv{CB}\\
& \iff \boxed{\vv{BF} = \vv{BA} + \vv{BC}}
\end{aligned}$
\end{itemize}

On a alors la figure suivante:

\begin{center}
\definecolor{ffqqqq}{rgb}{1,0,0}
\definecolor{qqzzqq}{rgb}{0,0.6,0}
\definecolor{qqwwzz}{rgb}{0,0.4,0.6}
\definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1}
\begin{tikzpicture}[>=latex]
\draw [->,line width=2pt,color=qqwwzz] (0,0) -- (4,0);
\draw [->,line width=2pt,color=qqzzqq] (0,0) -- (3,1);
\draw [->,line width=2pt,color=qqzzqq] (4,0) -- (7,1);
\draw [->,line width=2pt,color=qqzzqq] (7,1) -- (10,2);
\draw [->,line width=2pt] (4,0) -- (2.5,-0.5);
\draw [->,line width=2pt,color=ffqqqq] (4,0) -- (3,1);
\draw [->,line width=2pt,color=ffqqqq] (0,0) -- (-1,1);
\node at (-0.3554257869117754,-0.14746177279887482) {$A$};
\node at (3.9894555831044065,-0.535727767566278) {$B$};
\node at (3.138968165994856,1.257691351121251) {$C$};
\node at (10.146244928701805,2.2560896233802876) {$D$};
\node at (2.5288358885032216,-0.8130606209715661) {$E$};
\node at (-0.8546249230412941,1.4056022062707378) {$F$};
\end{tikzpicture}
\end{center}

\item Par construction, $\vv{BC} = \vv{AF}$. Ainsi, $ABCF$ est un parallélogramme.
\end{enumerate}
