Geometry problem from RMO 2016 The following problem is from RMO 2016. Initially it seems pretty trivial, but I am not able to find an easy or elegant solution. The official solution is not intuitive. I am looking for an alternate elegant proof, and also framed in a proper way like we do in contests, because in such problems showing the exact steps is very critical.

Let $ABC$ be a right-angled triangle with $\angle B=90^{\circ}$ degree. Let $I$ be the incentre of $ABC$. Let $AI$ extended intersect $BC$ in $F$. Draw a line perpendicular to $AI$ at $I$. Let it intersect $AC$ in $E$. Prove that $IE = IF$.

So far I have tried taking a point $E'$ such that $IE'=IF$ and then proving that $E$ and $E'$ coincide.

Observe that,  $\angle AEI=90-\frac {\angle A}{2}=\angle AFB$ and thereafter $\angle IEC=\angle IFC$.
In $\triangle IEC$ and $\triangle IFC$, $\angle ICE=\angle ICF$ and $\angle IEC=\angle IFC$ and therefore they are similar. Moreover,  corresponding side $CI$ is common to both of them so $\triangle IEC\cong \triangle IFC$ and hence $IE=IF$.
