JBMO-SL 2021 G2

Let $P$ be an interior point of the isosceles triangle $ABC$ with $\hat{A} = 90^{\circ}$ . If $$\widehat{PAB} + \widehat{PBC} + \widehat{PCA} = 90^{\circ},$$ prove that $AP \perp BC$ .

Proposed by Mehmet Akif Yıldız, Turkey