Let $K$ be a midpoint of $AH$. It is enough to prove that $MK=KD$. But this is evident, since $\angle KDM=\angle KMD=90^\circ-\frac{1}{2} \angle ACB$.

Alternatively, you can let $N$ be the midpoint of $ED$ and by angle chasing we can deduce $\angle MNA = 180^{\circ} - \angle C = \angle BHA$ and, since $M$ is the midpoint of $AB$, this implies $N$ is the midpoint of $AH$. This implies $AE = DH$.