IGO 2016 E2

Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$ . Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$ , such that $CX = CY = AB$ . (The points $A$ and $Y$ lie on different sides of the line $BC$ ). The line $XY$ intersects $\omega$ for the second time in point $P$ . Show that $PB = PC$ .

by Iman Maghsoudi