Auto: Zastassky A. A.

Two convex quadrilaterals are such that the sides of each lie on the perpendicular bisectors of the sides of the other. Find their angles.
Let the side $C^{\prime} D^{\prime}$ of the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ lie on the perpendicular bisector of the side $A B$ of the quadrilateral $A B C D$, and the side $D^{\prime} A^{\prime}$ on the perpendicular bisector of $B C$. Then $D^{\prime}$ is the center of the circumcircle of the triangle $A B C$. Similarly, $A^{\prime}, B^{\prime}, C^{\prime}$ are the centers of the circumcircles of the triangles $B C D, C D A, D A B$. Therefore, $B^{\prime} D^{\prime}$ is the perpendicular bisector of $A C$. In turn, $A C$ is the perpendicular bisector of one of the diagonals of the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and since the line $A C$ is perpendicular to $B^{\prime} D^{\prime}$ and $B^{\prime} D^{\prime}$ is not parallel to $A^{\prime} C^{\prime}$, $A C$ is the perpendicular bisector of $B^{\prime} D^{\prime}$, that is, $A B^{\prime} C D^{\prime}$ is a rhombus.

The composition of symmetries with respect to the lines $C^{\prime} D^{\prime}, D^{\prime} A^{\prime}, A^{\prime} B^{\prime}$, and $B^{\prime} C^{\prime}$ leaves the point $A$ in place and, therefore, is a rotation with center $A$. On the other hand, it is the composition of rotations with center $D^{\prime}$ by twice the angle $C^{\prime} D^{\prime} A^{\prime}$ and with center $B^{\prime}$ by twice the angle $A^{\prime} B^{\prime} C^{\prime}$, hence,

$\angle C^{\prime} B^{\prime} A^{\prime} = \angle A B^{\prime} D^{\prime} = \angle B^{\prime} D^{\prime} A = \angle A^{\prime} B^{\prime} C^{\prime}$. Similarly, $\angle B^{\prime} C^{\prime} D^{\prime} = \angle D^{\prime} A^{\prime} B^{\prime}$, that is, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a parallelogram. Since the sides of the quadrilateral $A B C D$ are perpendicular to the sides of the parallelogram $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, $A B C D$ is a parallelogram with the same angles.

Since $C$ is the center of the circumcircle of the triangle $B^{\prime} C^{\prime} D^{\prime}$, $\angle D^{\prime} C B^{\prime} = 2 \angle C^{\prime} D^{\prime} A^{\prime} = \angle B^{\prime} D^{\prime} C + \angle C B^{\prime} D^{\prime} = 90^{\circ}$. Accordingly, the acute angles of the parallelograms $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are $45^{\circ}$.

It is not difficult to see that two such parallelograms, obtained from each other by a $90^{\circ}$ rotation around a common center, satisfy the condition of the problem (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-10.jpg?height=821&width=840&top_left_y=1&top_left_x=615)

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