Problem 1353

Solution. Using that

$$\angle B B^{\prime} C = \angle C C^{\prime} B = 90^{\circ}, \quad \angle A A^{\prime} D = \angle D D^{\prime} A = 90^{\circ},$$

the quadrilaterals $B B^{\prime} C^{\prime} C$ and $A A^{\prime} D^{\prime} D$ are cyclic, just like $A B^{\prime} A^{\prime} B$ and $C D^{\prime} C^{\prime} D$.

By the inscribed angle theorem (in the cyclic quadrilateral $A B C D$), the chord $B C$ is seen from points $A$ and $D$ at the same angle, i.e., $\angle B A C = \angle B D C$. Similarly, in the cyclic quadrilateral $A A^{\prime} D^{\prime} D$, $\angle A^{\prime} A D^{\prime} = \angle A^{\prime} D D^{\prime}$. Thus,

$$\angle B A A^{\prime} = \angle B A C - \angle A^{\prime} A D^{\prime} = \angle B D C - \angle A^{\prime} D D^{\prime} = \angle D^{\prime} D C.$$

In the cyclic quadrilateral $A B A^{\prime} B^{\prime}$, $\angle A^{\prime} B^{\prime} B = \angle B A A^{\prime}$, and in the cyclic quadrilateral $C D C^{\prime} D^{\prime}$, $\angle D^{\prime} C^{\prime} C = \angle D^{\prime} D C$, so $\angle D^{\prime} C^{\prime} C = \angle A^{\prime} B^{\prime} B$, from which

$$\angle D^{\prime} C^{\prime} A^{\prime} = 90^{\circ} - \angle D^{\prime} C^{\prime} C = 90^{\circ} - \angle A^{\prime} B^{\prime} B = \angle A^{\prime} B^{\prime} D^{\prime},$$

thus, by the converse of the inscribed angle theorem, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a cyclic quadrilateral.

It remains to show that the center of the circle circumscribed around the cyclic quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is precisely the intersection of the diagonals of the quadrilateral $E F G H$. Since the lines $E F$ and $G H$ are perpendicular to the diagonal $A C$, and $E H$ and $F G$ are perpendicular to the diagonal $B D$, the quadrilateral $E F G H$ is a parallelogram, whose center is the intersection of its diagonals. The center of this parallelogram lies on the line through the midpoints of $E F$ and $G H$ (the midline of the parallelogram), which is perpendicular to $D^{\prime} B^{\prime}$, and every point on this line is equidistant from the lines $E F$ and $G H$.

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This implies that this line is the perpendicular bisector of the segment $D^{\prime} B^{\prime}$. Thus, we have shown that the center of the parallelogram $E F G H$ lies on the perpendicular bisector of the segment $D^{\prime} B^{\prime}$, and similarly, it lies on the perpendicular bisector of the segment $A^{\prime} C^{\prime}$. However, these two lines have only one common point, which is the center of the circle circumscribed around the cyclic quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$.

Remark. If the intersection of the diagonals is $M$, then without loss of generality, we can assume that the angle $\angle A M B$ is acute. In this case, the points $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ lie on the open rays $M B, M A, M D, M C$, respectively. For simplicity, we assumed that these points lie in the interiors of $M B, M A, M D, M C$; the statement of the problem can be proven similarly in other cases.