Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$. Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$. Bisector of $\angle BAC$ cuts $(K)$ again at $P$. The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$. The line $PK$ meets the perpendicular bisector of $BC$ at $F$. Prove that $\angle EGF = 90^o$.
Problem
Source: 2018 Saudi Arabia IMO TST IV p2
Tags: geometry, circumcircle, right angle, angle bisector
