KöMaL 2019 B5023

In a triangle $ABC$, $\angle ACB=90^{\circ}$ and $AC>BC$. Let $X$ be the midpoint of the arc $AB$ of the circumscribed circle that does not contain $C$. The perpendicular drawn to $CX$ at $X$ intersects line $CA$ at $P$. Show that $AP=BC$.