Denote $\omega$ the circumcircle of $\triangle ABC$ and $\gamma$ the circle tangent to $\omega$, $DA$, $DC$. Let $\omega$ touch $\gamma$ at $K$ and $M$ be the midpoint of $\operatorname{arc} BC$ 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 $F'$. Since $\omega$, $\gamma$ have the same tangent at $K$ then $\angle KF'I = \angle EF'K = \angle MAK = \angle KAI \Rightarrow AKIF'$ is cyclic. One has $\angle MCB = \angle MBC = \angle MKC \Rightarrow \triangle MEC \sim \triangle MCK$. Hence
$$MI^2 = MC^2 = ME \cdot MK \Rightarrow \triangle MEI \sim \triangle MIK .$$
Therefore, $\angle KEI = \angle AIK = \angle AF'K$. This implies that $AF'$ is tangent to $\gamma$, and $F = F'$.
Hence $E$, $I$, $F$ are collinear.
$\square$