11-7. Given a parallelogram $A B C D$, where $\angle B=111^{\circ}$ and $B C=B D$. On the segment $B C$, a point $H$ is marked such that $\angle B H D=90^{\circ}$. Point $M$ is the midpoint of side $A B$. Find the angle $A M H$. Give your answer in degrees.
Answer: $132^{\circ}$.

Solution. Note that $\angle D M B=90^{\circ}$, since $D A=D B$, and in the isosceles triangle $B D A$, the median $D M$ is also the altitude. Since angles $D H B$ and $D M B$ are right angles, points $M, B, H$, and $D$ lie on the same circle. It is clear that we need to find the angle $D M H$, as $\angle A M H=\angle D M H+90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-32.jpg?height=688&width=794&top_left_y=1398&top_left_x=631)

Angles $D M H$ and $D B H$ are equal, as these angles subtend the same arc $D H$ in the circle. Therefore, we need to find the angle $D B H$. This can be easily found from the isosceles triangle $D B C$ if we know the angle $C$ at the base of this triangle: $\angle D B C=180^{\circ}-2 \angle C$. But the angle $C$ is equal to $180^{\circ}-111^{\circ}=69^{\circ}$. Thus, $\angle D B C=42^{\circ}$, and $\angle A M H=\angle D M H+90^{\circ}=42^{\circ}+90^{\circ}=132^{\circ}$.