Let $BM$ be the median of the triangle $ABC$ with $AB > BC$. The point $P$ is chosen so that $AB\parallel PC$ and $PM \perp BM$. Prove that $\angle ABM = \angle MBP$. Proposed by Mykhailo Shandenko
Source: Kyiv City MO 2021 Round 1, Problem 8.4
Tags: geometry
Let $BM$ be the median of the triangle $ABC$ with $AB > BC$. The point $P$ is chosen so that $AB\parallel PC$ and $PM \perp BM$. Prove that $\angle ABM = \angle MBP$. Proposed by Mykhailo Shandenko