Question 65
Let $ABCD$ be an equilateral trapezoid with sides $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $\angle DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove that the triangle $AFG$ is isosceles.