1. Given a parallelogram $A B C D$. It is known that the centers of the circles circumscribed around triangles $A B C$ and $C D A$ lie on the diagonal $B D$. Find the angle $D B C$, if $\angle A B D=40^{\circ}$.
Answer: $50^{\circ}$ or $40^{\circ}$.

Solution. The center of the circle circumscribed around triangle $A B C$ lies on the perpendicular bisector of segment $A C$. Therefore, either this center is the midpoint of $A C$ (and then $A B C D$ is a rectangle), or $D B \perp A C$ (and then $A B C D$ is a rhombus). In the first case, angle $D B C$ complements angle $A B D$ to a right angle, and in the second case, these angles are equal to each other.

Evaluation. 10 points for a correct solution. If only the case of the rhombus is considered, 6 points. If only the case of the rectangle is considered, 4 points.