Croatian Mathematical Olympiad

Let $N$ be the other intersection of the circles circumscribed to the triangles $BMX$ and $CMY$.



Since $\angle BNM=\angle BXM=\angle CYM=\angle CNM$, the line $MN$ is the angle bisector of $\angle BNC$. The segment $NM$ is a median in the triangle $BNC$, hence $|BN|=|CN|$ and $NM\perp BC$, from which we conclude that $A,M$ and $N$ are collinear.

From $NM\perp BC$ we also have $\angle PXN=\angle BXN=180^{\circ }-\angle BMN=90^{\circ }$, and analogously $\angle PYN=90^{\circ }$, meaning that $P,X,N$ and $Y$ lie on the circle with diameter $PN$.

Finally, since $PA\parallel BC\perp AN$, we get $\angle PAN=90^{\circ }$, which completes the proof.