Baltic Way 1993 · Problem 19

Figure 7

Figure 8

Solution:

As the quadrangle $A B C D$ is inscribed in a circle, we have $\angle A B C + \angle C D A = \angle B C D + \angle D A B = 180^{\circ}$. It suffices to show that if each of these angles is equal to $90^{\circ}$, then each of the angles $A O B, B O C, C O D$ and $D O A$ is also equal to $90^{\circ}$ and thus $A B C D$ is a square. We consider the two possible situations:

(a) At least one of the diagonals of $A B C D$ is a diameter — say, $\angle A O B + \angle B O C = 180^{\circ}$. Then $\angle A B C = \angle C D A = 90^{\circ}$ and at least two of the angles $A O B, B O C, C O D$ and $D O A$ must be $90^{\circ}$: say, $\angle A O B = \angle B O C = 90^{\circ}$. Now, $\angle C O D = \angle D A B$ and $\angle D O A = \angle B C D$ (see Figure 7). Using the fact that $\frac{1}{2} \angle D O A = \angle D C A = \angle B C D - 45^{\circ}$ we have $\angle B C D = \angle D A B = 90^{\circ}$.

(b) None of the diagonals of the quadrangle $A B C D$ is a diameter. Then $\angle A O B + \angle C O D = \angle B O C + \angle D O A = 180^{\circ}$ and no angle of the quadrangle $A B C D$ is equal to $90^{\circ}$. Consequently, none of the angles $A O B, B O C, C O D$ and $D O A$ is equal to $90^{\circ}$. Without loss of generality we assume that $\angle A O B > 90^{\circ}$, $\angle B O C > 90^{\circ}$ (see Figure 8). Then $\angle A B C < 90^{\circ}$ and thus $\angle A B C = \angle C O D$ or $\angle A B C = \angle D O A$. As $\angle C O D + \angle D O A = \angle A O C = 2 \angle A B C$, we have $\angle C O D = \angle D O A$ and $\angle A O B + \angle D O A = 180^{\circ}$, a contradiction.