Balti Tee 2020 · Ülesanne 11

The key observation is that the circumcircle of $\triangle DPQ$ is tangent to $BC$. This can be proved by angle chasing:

$$\begin{aligned} \angle BDP &= 90^\circ - \angle PBD = 90^\circ - \angle PBC = 90^\circ - (180^\circ - \angle CQP) \\ &= \angle CQP - 90^\circ = \angle DQP. \end{aligned}$$

Now let the tangent to the circumcircle of $\triangle ABC$ at $A$ intersect $BC$ at $Y$. It is well-known (and easy to show) that $YA = YD$. This implies that $Y$ lies on the radical axis of the circumcircles of $\triangle ABC$ and $\triangle PDQ$, which is the line $PQ$. Thus $Y \equiv X$, and the claim follows.