Let $BM$ be the median of the triangle $ABC$, in which $AB> BC$. Point $P$ is chosen so that $AB \parallel PC$ and$ PM \perp BM$. The point $Q$ is chosen on the line $BP$ so that $\angle AQC = 90^o$, and the points $B$ and $Q$ lie on opposite sides of the line $AC$. Prove that $AB = BQ$. (Mikhail Standenko)