SHARYGIN 2015 Problem 3

The side $AD$ of a square $ABCD$ is the base of an obtuse-angled isosceles triangle $AED$ with vertex $E$ lying inside the square. Let $AF$ be a diameter of the circumcircle of this triangle, and $G$ be a point on $CD$ such that $CG = DF$. Prove that angle $BGE$ is less than half of angle $AED$.