Problem

Source: Poland 73-3-5

Tags: geometry, perpendicular bisector



Let $ABC$ be a triangle satisfying $AB<AC$. Let $M$ be the midpoint of $BC$. A point $P$ lies on the segment $AB$ with $AP>PB$. A point $Q$ lies on the segment $AC$ with $\angle MPA = \angle AQM$. The perpendicular bisectors of $BC$ and $PQ$ intersect at $S$. Prove that $\angle BAC + \angle QSP = \angle QMP$.