Problem

Source: China TST Test 2 Day 2 Q5

Tags: geometry, China TST



Let $M$ be the midpoint of $BC$ of triangle $ABC$. The circle with diameter $BC$, $\omega$, meets $AB,AC$ at $D,E$ respectively. $P$ lies inside $\triangle ABC$ such that $\angle PBA=\angle PAC, \angle PCA=\angle PAB$, and $2PM\cdot DE=BC^2$. Point $X$ lies outside $\omega$ such that $XM\parallel AP$, and $\frac{XB}{XC}=\frac{AB}{AC}$. Prove that $\angle BXC +\angle BAC=90^{\circ}$.