The triangle $ABC$ is right-angled at $A$. Its incentre is $I$, and $H$ is the foot of the perpendicular from $I$ on $AB$. The perpendicular from $H$ on $BC$ meets $BC$ at $E$, and it meets the bisector of $\angle ABC$ at $D$. The perpendicular from $A$ on $BC$ meets $BC$ at $F$. Prove that $\angle EFD = 45^o$
Problem
Source: Irmo 2018 p1 q2
Tags: geometry, right triangle, Angle Chasing
TheDarkPrince
16.09.2018 14:30
Let the feet of $D$ on $AB$ be $G$. We have \[\frac{BF}{AB} = \frac{EF}{AH} \implies EF = \frac{BF\cdot AH}{AB}.\]Also, \begin{align*} ED = GD &= \frac{IH\cdot BG}{BH} = \frac{IH\cdot BE}{BH} \\\\ &= \frac{IH\cdot BF}{BA} = \frac{AH\cdot BF}{AB} = EF. \end{align*}Thus $EF = ED$ or $EDF$ is isosceles right triangle and we are done.
sunken rock
16.09.2018 23:13
Easily $\angle DHI=\angle ABC$; with $\angle DIH=90^\circ-\frac{\angle ABC}2$ we infer $DH=HI(=AH)$, that is, $H$ is circumcenter of $\triangle ADI\implies \angle ADI=45^\circ$, or $D$ is incenter of $\triangle ABF$, done. Best regards, sunken rock
AopsUser101
17.03.2020 01:04
I did the problem on my own, and it resembles sunken rock’s solution. I’ll write it here anyways: Assume WLOG $AB<AC$. We have that $\angle IAC = \angle HAI = \angle HIA = 45$. Let $\angle IBH = \angle IBC=x$. Then, $\angle IBH = 90-x \implies \angle BHE = 90-2x \implies \angle DHI = 2x \implies \angle HDI = 90-x$. Since $\Delta ADI$ as $HI=HA$ and $HD=HI$, $HA=HD$ and $\angle HDA = \angle HAD = 45-x$. As a result, $D$ is the incenter of $AFB$, proving that $\angle DFE = 45$ .