\begin{enumerate}
\item On a la figure suivante :

\medskip

\begin{minipage}{5cm}
\begin{center}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (3,0);
\coordinate (C) at (1,2);
\coordinate (E) at ($0.6*(A)+0.4*(C)$);
\coordinate (F) at ($0.25*(A)+0.75*(B)$);
\coordinate (H) at (${9/7}*(B)-{2/7}*(C)$);
\draw (A) node[left] {A} -- (B) node[right] {B} -- (C) node[above] {C} -- cycle;
\draw[dashed] (B) -- (H);
\fill[red] (E) circle (1pt) node[left] {E};
\fill[red] (F) circle (1pt) node[below] {F};
\fill[red] (H) circle (1pt) node[right] {H};
%
\fill ($0.2*(A)+0.8*(C)$) circle (1pt);
\fill ($0.4*(A)+0.6*(C)$) circle (1pt);
\fill ($0.8*(A)+0.2*(C)$) circle (1pt);
\fill ($0.75*(A)+0.25*(B)$) circle (1pt);
\fill ($0.5*(A)+0.5*(B)$) circle (1pt);
\fill (${1/7}*(B)+{6/7}*(C)$) circle (1pt);
\fill (${2/7}*(B)+{5/7}*(C)$) circle (1pt);
\fill (${3/7}*(B)+{4/7}*(C)$) circle (1pt);
\fill (${4/7}*(B)+{3/7}*(C)$) circle (1pt);
\fill (${5/7}*(B)+{2/7}*(C)$) circle (1pt);
\fill (${6/7}*(B)+{1/7}*(C)$) circle (1pt);
\fill (${8/7}*(B)-{1/7}*(C)$) circle (1pt);
\end{tikzpicture}
\end{center}
\end{minipage}
\begin{minipage}{\dimexpr\linewidth-5.1cm}
\begin{itemize}
\item $\vv{\text{EC}}=\dfrac{3}{5}\vv{\text{AC}}$ donc $\vv{\text{CE}}=\dfrac{3}{5}\vv{\text{CA}}$ (il est plus simple de partir d'un point que l"'on conna\^it (ici, le point C) pour construire le point E.

\item $\vv{\text{CH}}=-\dfrac{9}{7}\vv{\text{BC}}$ donc $\vv{\text{CH}}=\dfrac{9}{7}\vv{\text{CB}}=\vv{\text{CB}}+\dfrac{2}{7}\vv{\text{CB}}$.
\end{itemize}
\end{minipage}

\item $\begin{aligned}[t]
\vv{\text{EF}} & = \vv{\text{EC}}+\vv{\text{CA}}+\vv{\text{AB}}+\vv{\text{BF}}\\
& = \dfrac{3}{5}\vv{\text{AC}}-\vv{\text{AC}}+\vv{\text{AB}}-\dfrac{1}{4}\vv{\text{AB}}\\
\Aboxed{\vv{\text{EF}} & = \dfrac{3}{4}\vv{\text{AB}}-\dfrac{2}{5}\vv{\text{AC}}}
\end{aligned}$

\item $\begin{aligned}[t]
\vv{\text{EH}} & = \vv{\text{EC}}+\vv{\text{CH}}\\
& = \dfrac{3}{5}\vv{\text{AC}}+\dfrac{9}{7}\vv{\text{CB}}\\
& = \dfrac{3}{5}\vv{\text{AC}}+\dfrac{9}{7}\left(\vv{\text{CA}}+\vv{\text{AB}}\right)\\
& = \dfrac{3}{5}\vv{\text{AC}}-\dfrac{9}{7}\vv{\text{AC}}+\dfrac{9}{7}\vv{\text{AB}}\\
\Aboxed{\vv{\text{EH}}& = \dfrac{9}{7}\vv{\text{AB}}-\dfrac{24}{35}\vv{\text{AC}}}
\end{aligned}$

\item Dans le rep�re $\big(\text{A}~;~\vv{\text{AB}}~;~\vv{\text{AC}}\big)$, on a :
\[
\vv{\text{EF}}\dbinom{\frac{3}{4}}{-\frac{2}{5}}\qquad\text{et}\qquad\vv{\text{EH}}\dbinom{\frac{9}{7}}{-\frac{24}{35}}.
\]
\[
\dfrac{3}{4}\times\left(-\dfrac{24}{35}\right)-\dfrac{9}{7}\times\left(-\dfrac{2}{5}\right)=-\dfrac{18}{35}+\dfrac{18}{35}=0.
\]

Ainsi, $\vv{\text{EF}}$ et $\vv{\text{EH}}$ sont colin�aires ; ils ont un point en commun (E), donc les points E, F et H sont align�s.
\end{enumerate}