\begin{enumerate}
\item La figure est la suivante :
\begin{center}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (5,0);
\coordinate (C) at (7,3);
\coordinate (D) at (2,3);
\coordinate (H) at (${2/3}*(A)+{1/3}*(B)$);
\coordinate (K) at (${1/3}*(A)+{2/3}*(D)$);
\coordinate (I) at (${1/2}*(B)+{1/2}*(C)$);
\coordinate (J) at (${1/2}*(D)+{1/2}*(C)$);
\draw (A) node[left] {A} -- (B) node[right] {B} -- (C) node[right] {C} -- (D) node[left] {D} -- cycle;
\fill (H) circle (1pt) node[below] {H};
\fill (K) circle (1pt) node[left] {K};
\fill (I) circle (1pt) node[right] {I};
\fill (J) circle (1pt) node[above] {J};
\fill (${1/3}*(A)+{2/3}*(B)$) circle (1pt);
\fill (${2/3}*(A)+{1/3}*(D)$) circle (1pt);
\end{tikzpicture}
\end{center}

\item $\begin{aligned}[t]
\vv{\text{HI}} & = \vv{\text{HB}}+\vv{\text{BI}}\\
& = \dfrac{1}{3}\vv{\text{AB}}+\dfrac{1}{2}\vv{\text{BC}}\\
\Aboxed{\vv{\text{HI}}& = \dfrac{2}{3}\vv{\text{AB}}+\dfrac{1}{2}\vv{\text{AD}}}
\end{aligned} $

\medskip

$\begin{aligned}[t]
\vv{\text{KJ}} & = \vv{\text{KD}}+\vv{\text{DJ}}\\
& = \dfrac{1}{3}\vv{\text{AD}}+\dfrac{1}{2}\vv{\text{DC}}\\
\Aboxed{\vv{\text{KJ}} & = \dfrac{1}{2}\vv{\text{AB}}+\dfrac{1}{3}\vv{\text{AD}}}
\end{aligned} $

\item Dans le repère $\big(\text{A}~;~\vv{\text{AB}}~;~\vv{\text{AD}}\big)$, on a :
\[
\vv{\text{HI}}
\dbinom{\frac{2}{3}}{\frac{1}{2}}\qquad\text{et}\qquad\vv{\text{KJ}}
\dbinom{\frac{1}{2}}{\frac{1}{3}}.
\]
\[
\dfrac{2}{3}\times\dfrac{1}{3}-\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{2}{9}-\dfrac{1}{4}\neq0.
\]

Ainsi, les vecteurs $\vv{\text{HI}}$ et $\vv{\text{KJ}}$ ne sont pas colinéaires.

\medskip

\underline{Les droites (HI) et (KJ) ne sont donc pas parallèles.}
\end{enumerate}
\newpage