Question 69
Let $ABCD$ be a cyclic quadrangle. The perpendicular bisector of $\overline{BC}$ meets the side $\overline{AB}$ in point $E$. The circle through $E$, $C$ and the midpoint $F$ of the side $\overline{BC}$ meets the side $\overline{CD}$ in $G$. Prove that $AD \perp FG$. (D. Monk)