Answer: $90^{\circ}$. Solution: Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, $E N \perp A D$. Since $A B \| M N$ and $\angle A B E=90^{\circ}$, then $B E \perp M N$. Therefore, $E$ is the orthocenter of triangle $B M N$. This means $M E \perp B N$. Since $B M D N$ is a parallelogram, $B N \| D M$, hence $\angle D M E=90^{\circ}$.

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4. Inside parallelogram $A B C D$, a point $E$ is chosen such that $A E=D E$ and $\angle A B E=90^{\circ}$. Point $M$ is the midpoint of segment $BC$. Find the angle $DME$. (A. Kuznetsov)