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.


You can solve the problem only playing with angles. In the picture, all red angles are equal to $90^{\circ}$. In addition, we will prove that all green angles are the same, so $\angle BAF=\angle FAC=\angle GIF=\angle EIH$.
That $\angle BAF=\angle FAC$ is obvious because $AF$ is a bissection. $IG$ is paralel to $AB$, so $\angle BAF=\angle GIF$. In addition, because $\angle EIH=90^{\circ}$ then $\angle EIH=\angle FAC$.
Finally, $GI=IH$ because both are in-radius. Therefore, the triangles $GIF$ and $HIE$ are congruent by the case $ASA$. It implies that $IF=IE$.
