Question 65
Let $\omega$ be the circumcircle of an acute-angled triangle $ABC$. Point $D$ lies on the arc $BC$ of $\omega$ not containing point $A$. Point $E$ lies in the interior of the triangle $ABC$, does not lie on the line $AD$, and satisfies $\angle DBE = \angle ACB$ and $\angle DCE = \angle ABC$. Let $F$ be a point on the line $AD$ such that lines $EF$ and $BC$ are parallel, and let $G$ be a point on $\omega$ different from $A$ such that $AF = FG$. Prove that points $D, E, F, G$ lie on one circle.
(Slovakia)
(Slovakia)