Solution: Let $O, K, L$ be the midpoints of $BC, AC, BD$, and $P$ be the intersection point of $AC$ and $BD$. Points $K$ and $L$ are distinct (otherwise $ABCD$ is a rhombus and $\left.\angle BMC < \angle BPC = 90^{\circ}\right)$.

![](https://cdn.mathpix.com/cropped/2024_06_03_62c72dea8e282f2d2f19g-4.jpg?height=334&width=1151&top_left_y=1049&top_left_x=424)

Triangles $\triangle ABC$ and $\triangle BCD$ are isosceles, so the median to the base in each of them is also an altitude, i.e., $BK \perp AC$ and $CL \perp BD$. From the equality

$$
\angle BKC = \angle BMC = \angle BLC = 90^{\circ}
$$

it follows that points $K, M, L$ lie on a circle with diameter $BC$ and center point $O$. Since the chord $KM$ is the midsegment in $\triangle ACD$, we get that $KM = \frac{1}{2} CD = OC$, i.e., this chord is equal to the radius. Then $\triangle KOM$ is equilateral and $\angle MOK = 60^{\circ}$.

Similarly, $\angle MOL = 60^{\circ}$, i.e., $\angle KOL = 120^{\circ}$. The inscribed angle $\angle KBL$ is measured by the arc $\widehat{KL}$ (or by its complement when the given quadrilateral is not convex), so it is equal to $60^{\circ}$ (or $120^{\circ}$). This means that $\angle PBK = 60^{\circ}$, i.e., $\angle BPK = 30^{\circ}$.
Problem 6. In the quadrilateral $A B C D$, the sides $A B, B C$ and $C D$ are equal, and point $M$ is the midpoint of $A D$. It is known that $\angle B M C=90^{\circ}$. Find the angle between the diagonals of the quadrilateral $A B C D$.