Question 14
Let $\Gamma$ be the circumcircle of acute triangle $ABC$. The perpendicular to $AB$ from $C$ meets $AB$ at $D$ and $\Gamma$ again at $E$. The bisector of angle $C$ meets $AB$ at $F$ and $\Gamma$ again at $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.
