Let $BM$ be the median of triangle $ABC$ in which $AB > BC$. The point $P$ is chosen so that $AB\parallel PC$ and $PM \perp BM$. On the line $BP$, point $Q$ is chosen so that $\angle AQC = 90^\circ$, and points $B$ and $Q$ are on opposite sides of the line $AC$. Prove that $AB = BQ$. Proposed by Mykhailo Shtandenko