
\begin{itemize}
\item $\begin{aligned}[t]
4\vv{AM}-5\vv{MB}=\vv{0} & \iff 4\vv{AM}-5\big(\vv{MA}+\vv{AB}\big)=\vv{0}\\
& \iff 4\vv{AM}-5\vv{MA}-5\vv{AB}=\vv{0}\\
& \iff 4\vv{AM}+5\vv{AM}-5\vv{AB}=\vv{0}\\
& \iff 9\vv{AM}=5\vv{AB}.
\end{aligned}$

\item $\begin{aligned}[t]
9\vv{BN}+4\vv{AB}-5\vv{BC}=\vv{0} & \iff 9\big(\vv{BA}+\vv{AN}\big)+4\vv{AB}-5\big(\vv{BA}+\vv{AC}\big)=\vv{0}\\
& \iff 9\vv{BA}+9\vv{AN}+4\vv{AB}-5\vv{BA}-5\vv{AC}=\vv{0}\\
& \iff  -9\vv{AB}+9\vv{AN}+4\vv{AB}+5\vv{AB}-5\vv{AC}=\vv{0}\\
& \iff 9\vv{AN} = 5\vv{AC}.
\end{aligned}$
\end{itemize}

Donc :

\begin{align*}
9\vv{MN} & = 9\vv{MA} + 9\vv{AN}\\
        & = 5\vv{BA} + 5\vv{AC}\\
        & = 5\vv{BC}.
\end{align*}

Ainsi, $\vv{MN}$ et $\vv{BC}$ sont colinéaires. Par conséquent, $(MN)$ et $(BC)$ sont parallèles.
