Let $P$ be the foot of perpendicular from $A$ to $BC$. It follows from $\angle ADC = \angle APC = 90^\circ$ that $A$, $D$, $P$, $C$ are concyclic. Considering the power of $M$, we find that
$$ME \times MB = MD \times MA = MP \times MC.$$
Since $MB = MC$, we have $ME = MP$. Note that $FM$ is the perpendicular bisector of $BC$. Therefore, $FM \parallel AP$. As $ME = MP$, it follows from the intercept theorem that $F$ is the midpoint of $AE$.