The angle at vertex $A$ of the rhombus $A B C D$ is $20^{\circ}$. Points $M$ and $N$ are the feet of the perpendiculars dropped from vertex $B$ to sides $A D$ and $C D$.

Find the angles of triangle $B M N$.
Prove that triangle $M B N$ is isosceles.

## Solution

From the equality of right triangles $A B M$ and $C B N$, it follows that $B=B N, \angle A B M=\angle C B N=70^{\circ}$. Therefore, $\angle M B N=\angle A B C-2 \angle A B M=20^{\circ}$, and since triangle $M B N$ is isosceles, $\angle B M N=\angle B N M=1 / 2\left(180^{\circ}-\right.$ $\left.20^{\circ}\right)=80^{\circ}$.

## Answer

$20^{\circ}, 80^{\circ}, 80^{\circ}$.