Problem

Source: CSEMO 2018 Grade 11 Q3

Tags: geometry, parallelogram



Let $O$ be the circumcenter of $\triangle ABC$, where $\angle ABC> 90^{\circ}$ and $M$ is the midpoint of $BC$. Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$. $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$. If quadrilateral $ADOE$ is a parallelogram, prove that $$\angle OPE = \angle AMB.$$