Answer
See explanation.
Work Step by Step
Let $B\left(b_{1}, b_{2}\right), C\left(c_{1}, c_{2}\right), and A\left(a_{1}, a_{2}\right)$ be the vertecies of the triangle. $M_{A B}=\left(\frac{b_{2}+a_{2}}{2}, \frac{b_{1}+a_{1}}{2}\right)$ is the middle point of the segment $A \vec{B}$
$M_{A C}=\left(\frac{c_{1}+a_{1}}{2}, \frac{c_{2}+a_{2}}{2}\right)$ is the middle point of the segment $A C$
Let $\mathbf{u}$ be the vector $\mathbf{u}= \vec{M}_{A C} M_{A B}$
\[
\begin{aligned}
\vec{u}=\left\{\frac{a_{1}+c_{1}}{2}-\frac{b_{2}+a_{2}}{2}, \frac{a_{2}+c_{2}}{2}-\frac{b_{1}+a_{1}}{2}\right.&) \\
&=\left[\left\langle c_{1} c_{2}\right)-\left\langle b_{1}, b_{2}\right\rangle\right]\frac{1}{2}=\frac{1}{2} \vec{v}
\end{aligned}
\]
where $\vec{v}$ is the vector $\overrightarrow{B C}$.
Since $\frac{1}{2} \vec{v}=\vec{u},$ they are parallel vectors.
