SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $\omega$ the circumcircle of $△ABC$ and $\gamma$ the circle tangent to $\omega$, $DA$, $DC$. Let $\omega$ touch $\gamma$ at $K$ and $M$ be the midpoint of $\mathrm{arc}BC$ on $\omega$ not containing $K$. One has the dilation with center $K$ sending $\gamma$ to $\omega$ and $BC$ to the tangent of $\omega$ at $M$. Hence $K$, $E$, $M$ are collinear. We also have $A$, $I$, $M$ are collinear and $MI=MC$.

Let $EI$ meet $\gamma$ again at $F^{\prime }$. Since $\omega$, $\gamma$ have the same tangent at $K$ then $\angle K{F}^{\prime }I=\angle E{F}^{\prime }K=\angle MAK=\angle KAI\Rightarrow AKI{F}^{\prime }$ is cyclic. One has $\angle MCB=\angle MBC=\angle MKC\Rightarrow △MEC\sim △MCK$. Hence $$M{I}^2=M{C}^2=ME\cdot MK\Rightarrow △MEI\sim △MIK.$$ Therefore, $\angle KEI=\angle AIK=\angle A{F}^{\prime }K$. This implies that $A{F}^{\prime }$ is tangent to $\gamma$, and $F=F^{\prime }$. Hence $E$, $I$, $F$ are collinear. $\square$