CN-TST 2019 Problem 11

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 $\angle BXC +\angle BAC=90^{\circ}$ .