Let $S$ be the midpoint of $BF$ and let $T$ be the midpoint of segment $AF$. We will prove that $\triangle MST \cong \triangle EDM$. $AB = CD \Rightarrow AD\|BC$ and$ \angle BAD = \angle CDA = \alpha $. $ED = BC \Rightarrow DC \| BE$. Now $DFBC$ is a parallelogram $\Rightarrow \angle CBF = \angle ADC = \angle DFE = \angle DEF = \alpha$. Now we have that $DC = FB = AB$ and $DE=DF=BC$. Observe that $TS = \frac{1}{2}AB = \frac{1}{2}DC = DM$ and $MS =DF=DE$. It remains to show that $\angle TSM = \angle EDM$. Indeed, $\angle EDM = 180^{\circ} - \alpha$ and $\angle TSM = \angle TSF+ \angle FSM = 180^{\circ}-2\alpha+\alpha = 180^{\circ}-\alpha$ and we are done.

Let $P$ be the midpoint of $AF$. Let $BM$ and $AD$ intersect at point $K$. Then, $BCKD$ is a parallelogram, and so $BM=MK$. Moreoever, an easy angle chase yields that triangle $ABF$ is isosceles, and so $BP$ is perpendicular to $AF$, that is triangle $BPK$ is right. Therefore, $MP=MB=MK$. To finish, $MB=ME$ due to symmetry, and so we have $MP=ME,$ as desired.