9.3. 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 $B C$. Find the angle $D M E$.
Answer: $90^{\circ}$.

First solution. Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, its median $E N$ is also an altitude, that is, $E N \perp A D$. Therefore, $N E \perp B C$ (see Fig. 2).

Since $A D \| B C$ and $B M = M C = A N = N D = A D / 2$, quadrilaterals $A B M N$ and $B M D N$ are parallelograms, from which $A B \| M N$ and $B N \| D M$. Since $\angle A B E = 90^{\circ}$ and $A B \| M N$, we get $B E \perp M N$. Thus, $E$ is the orthocenter of triangle $B M N$, from which $M E \perp B N$. Finally, from $B N \| D M$, we get $M E \perp D M$, that is, $\angle D M E = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfd41867062ad51c6492g-4.jpg?height=345&width=463&top_left_y=663&top_left_x=269)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_bfd41867062ad51c6492g-4.jpg?height=355&width=424&top_left_y=658&top_left_x=803)

Fig. 3

Second solution. On the extension of segment $A B$ beyond point $B$, mark point $F$ such that $A B = B F$ (see Fig. 3). Then $B F = C D$ and $B F \| C D$, so quadrilateral $B D C F$ is a parallelogram. Point $M$ is the midpoint of diagonal $B C$ of this parallelogram, hence $M$ is also the midpoint of its diagonal $D F$.

Line $B E$ is the perpendicular bisector of segment $A F$, from which $A E = F E$. From the condition, we now get that $D E = A E = F E$. Therefore, point $E$ lies on the perpendicular bisector of $D F$. Hence, $\angle E M D = 90^{\circ}$.