Prove that $MN$ is parallel to $KL$.

If you do not want to use coordinates, you can prove it by picture. The picture below shows you two solutions.

You can construct point $P$ so that $PF$ is parallel to $AB$ and $PE$ is parallel to $AC$. Then $AEPF$ is a parallelogram and then one can easily see that triangle $FPD$ is congruent to triangle $BMN$ as well as triangle $EPD$ is congruent to triangle $CKL$. Consequently, $\angle \, PDN = \angle \, MND $ and $\angle \, PDL = \angle \, KLD$ so $MN$ is parallel and equal to $PD$ and $KL$ is parallel and equal to $PD$. Thus $MN$ is parallel and equal to $KL$.

The other solutions is similar, but you construct the point $Q$ instead. Again you end up with congruent triangles and analogous arguments. Indeed, construct point $Q$ so that $QE$ is parallel to $BF$ and $QF$ is parallel to $CE$. Then $DEQF$ is a parallelogram and then one can easily see that triangle $EAQ$ is congruent to triangle $BMN$ as well as triangle $FAQ$ is congruent to triangle $CKL$. Consequently, $\angle \, BMN = \angle \, EAQ $ and $\angle \, CKL = \angle \, FAQ$ so $MN$ is parallel and equal to $AQ$ and $KL$ is parallel and equal to $AQ$. Thus $MN$ is parallel and equal to $KL$.

The vectors will do:

\begin{eqnarray*} \overrightarrow{NM} &=& \overrightarrow{NB}+\overrightarrow{BM}\\ &=& \overrightarrow{FD}+\overrightarrow{EA}\\ &=& \overrightarrow{D}-\overrightarrow{F}+\overrightarrow{A}-\overrightarrow{E} \end{eqnarray*} \begin{eqnarray*} \overrightarrow{LK} &=& \overrightarrow{LC} + \overrightarrow{CK}\\ &=& \overrightarrow{ED} + \overrightarrow{FA}\\ &=& \overrightarrow{D}-\overrightarrow{E}+\overrightarrow{A}-\overrightarrow{F} \end{eqnarray*} Since $\overrightarrow{NM}=\overrightarrow{LK}$ the quadrilateral $MNLK$ is a parallelogram.

Since affine maps preserve midpoints and parallel lines, it is not restrictive to assume that $A$ lies at the origin, $B$ has coordinates $(1,0)$, $C$ has coordinates $(0,1)$, $F$ has coordinates $(0,c)$, $E$ has coordinates $(b,0)$. Find the coordinates of $D,M,K,N,L$ and the claim will be straightforward to prove. An equivalent approach is to use barycentric coordinates.