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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.

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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 $IF=IE$.

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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 $IE=IF$.

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Extend $IE$ to meet the line $AB$ at $X$. Clearly, $\Delta AXE $ must be isosceles. Now, drop perpendiculars $PI$ and $P'I$ on $AB$ and $BC$.Thus,$PI=P'I$. We further see that $\Delta IP'F\cong \Delta IPX$ and thus $IX=IF$ and hence $IE=IF$.

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Extend IE from I to intersect the extension pBC at E'. Draw altitudes of triangles AIE and AIE' from I. Theses altitudes are equal because I is on the bisector of angle ACB. So two right angle triangles AIE and AIE' are equal because their altitudes are equal , so is their sides so $IE=IF$.

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Drop perpendiculars $IG,IH,IJ$ on $AB,BC,AC$ respectively. Let $AB=c$ and the inradius be $r$. Then $\triangle AIG \sim \triangle IFH$ and so $$FH = \frac{r^2}{c-r} $$ Similarly, by considering $\triangle AJI$ and $\triangle IJE$, $$EJ =\frac{r^2}{c-r}$$ So, $FH=EJ$ and $IH=IJ$ implying that $$\triangle IHF \cong \triangle IJE \implies IF=IE$$

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