Auto: Witness $M$.

In quadrilateral $A B C D$, sides $A B, B C$ and $C D$ are equal, $M$ is the midpoint of side $A D$. It is known that $\angle B M C=90^{\circ}$. Find the angle between the diagonals of quadrilateral $A B C D$.
Let $O, K, L$ be the midpoints of segments $B C, A C$ and $B D$ respectively, $P$ be the intersection point of lines $A C$ and $B D$. Points $K, L$ are distinct (otherwise $A B C D$ is a rhombus and

$\angle B M C<\angle B P C=90^{\circ}$). Since angles $B K C, B M C$ and $B L C$ are right (medians in isosceles triangles are altitudes), points $K, M, L$ lie on a circle with diameter $B C$. Chord $K M=1 / 2 C D=O C$ as the midline of triangle $A C D$, so triangle $K O M$ is equilateral and $\angle M O K=60^{\circ}$. Similarly, $\angle M O L=60^{\circ}$, so $\angle K O L=120^{\circ}$. The inscribed angle $K B L$ subtends arc $K L$ or its complement, so it equals $60^{\circ}$ or $120^{\circ}$. In any case, this means that in the right triangle $B K P$ the angle $B$ is $60^{\circ}$, so $\angle B P K=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c45034c1b1e9c9ba4d1bg-3.jpg?height=686&width=598&top_left_y=237&top_left_x=730)

## Answer

$30^{\circ}$.