# Geometry problem from RMO 2016 [closed]

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 $$IF=IE$$.

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 $$IE=IF$$.

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 $$IE=IF$$.

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 $$IE=IF$$.

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 $$\triangle IHF \cong \triangle IJE \implies IF=IE$$