Question 128
Let $ABC$ be a scalene triangle inscribed in circle $\Gamma$. The internal bisector of $\angle A$ meets $\overline{BC}$ and circle $\Gamma$ at points $D$ and $E$. The circle with diameter $\overline{DE}$ meets $\Gamma$ at a second point $F$. Prove that $\frac{AB}{AC}=\frac{FB}{FC}$.