BALKAN-SL 2009 G4

Let $MN$ be a line parallel to the side $BC$ of a triangle $ABC$, with $M$ on the side $AB$ and $N$ on the side $AC$. The lines $BN$ and $CM$ meet at point $P$. The circumcircles of triangles $BMP$ and $CNP$ meet at two distinct points $P$ and $Q$. Prove that $\angle BAQ = \angle CAP$.

Liubomir Chiriac, Moldova