Question 11
Let $ABC$ be a triangle with points $D$ and $E$ taken on side $BC$ such that $D$ is between $B$ and $E$. A point $F$ lies on segment $AD$ and satisfies the condition $\frac{AF}{FD} = \frac{BE}{EC}$. The circumcircle of triangle $ABD$ intersects segment $AE$ at a point $G$ (distinct from $A$). If lines $EF$ and $GC$ are parallel, prove that $GE = EC$.