Question 60
Let $ABC$ be a triangle, and let $M$ be a point on the side $AC$, different from both $A$ and $C$. The line through $M$ and parallel to $BC$ crosses the side $AB$ at $N$. The segments $BM$ and $CN$ cross at $P$, and the circles $BNP$ and $CMP$ meet again at $Q$. Show that the angles $BAP$ and $CAQ$ are equal.