, 2016

For every point $D$ on the side $BC$, let $D'$ be the reflection of $D$ in the line $PQ$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.

Choose $D$ to be the point where the angle bisector from $A$ meets $BC$. Note that $P$ and $Q$ lie on the rays $AB$ and $AC$ respectively. Furthermore, $P$ and $Q$ are reflections of each other in the line $AD$, from which it follows that $PQ \perp AD$. Therefore, $D'$ lies on the line $AD$ and we may deduce that either $D' = A$ or $D'$ is the second point of the angle bisector at $A$ and the circumcircle of $ABC$. However, since $APDQ$ is a cyclic quadrilateral, the segment $PQ$ intersects the segment $AD$. Therefore, $D'$ lies on the ray $DA$ and therefore $D' = A$. By angle chasing we obtain
$$\angle PD'Q = \angle PDQ = 180^{\circ} - \angle BAC$$
and since $D' = A$ we also know $\angle PD'Q = \angle BAC$. This implies that $\angle BAC = 90^{\circ}$.

Now we choose $D$ to be the midpoint of $BC$. Since $\angle BAC = 90^{\circ}$, we can deduce that $DQP$ is the medial triangle of triangle $ABC$. Therefore, $PQ \parallel BC$ from which it follows that $DD' \perp BC$. But the distance from $D'$ to $BC$ is equal to both the circumradius of triangle $ABC$ and to the distance from $A$ to $BC$. This can only happen if $A = D'$. This implies that $ABC$ is isosceles and right-angled at $A$.

We will now prove that if $ABC$ is isosceles and right-angled at $A$ then the required property in the problem holds. Let $D$ be any point on side $BC$. Then $D'P = DP$ and we also have $DP = BP$. Hence, $D'P = BP$ and similarly $D'Q = CQ$. Note that $APDQD'$ is cyclic with diameter $PQ$. Therefore, $\angle APD' = \angle AQD'$, from which we obtain $\angle BPD' = \angle CQD'$. So triangles $D'PB$ and $D'QC$ are similar. It follows that $\angle PD'Q = \angle PD'C + \angle CD'Q = \angle PD'C + \angle BD'P = \angle BD'C$ and $\frac{D'P}{D'Q} = \frac{D'B}{D'C}$. So we also obtain that triangles $D'PQ$ and $D'BC$ are similar. But since $DPQ$ and $D'PQ$ are congruent, we may deduce that $\angle BD'C = \angle PD'Q = \angle PDQ = 90^{\circ}$. Therefore, $D'$ lies on the circle with diameter $BC$, which is the circumcircle of triangle $ABC$.