Let $P, Q$ be points taken on the side $BC$ of a triangle $ABC$, in the order $B, P, Q, C$. Let the circumcircles of $\vartriangle PAB$, $\vartriangle QAC$ intersect at $M$ ($\ne A$) and those of $\vartriangle PAC, \vartriangle QAB$ at N. Prove that $A, M, N$ are collinear if and only if $P$ and $Q$ are symmetric in the midpoint $A' $ of $BC$.