The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.
Problem
Source: Sharygin Geometry Olympiad, Final Round 2016, Problem 1 grade 9
Tags: geometry, Angle Chasing, parallelogram, circumcircle
