Problem

Source: I Caucasus 2015 9.3

Tags: geometry, equal angles, equal segments, circumcircle, angle bisector



Let $AL$ be the angle bisector of the acute-angled triangle $ABC$. and $\omega$ be the circle circumscribed about it. Denote by $P$ the intersection point of the extension of the altitude $BH$ of the triangle $ABC$ with the circle $\omega$ . Prove that if $\angle BLA= \angle BAC$, then $BP = CP$.