Let $N$ be the midpoint of $AB$. Then $MN,NR$ are midlines in triangles $APB,ABC$ respectively. So, $MN=NR=\frac{AC}{2} \longrightarrow \triangle NMR$ is isosceles. Let the projection from $E$ onto $MR$ be $F$. Let the projection from $N$ to $MR$ be $G$. So $NG \parallel EF$. But $NG$ is an angle bisector. So, $\angle MNG = \angle BEF$ since $BE \parallel MN$ and $EF \parallel NG$. Furthermore, $\angle RNG =\angle AEF$ since $RN \parallel AC$ and $EF \parallel NG$. But $\angle MNG = \angle RNG$ implying $\angle AEF = \angle BEF$. We are done.