Let ${ABC}$ be a triangle and ${\Gamma}$ the circle with diameter ${AB}$. The bisectors of ${\angle BAC}$ and ${\angle ABC}$ intersect ${\Gamma}$ (also) at ${D}$ and ${E}$, respectively. The incircle of ${ABC}$ meets ${BC}$ and ${AC}$ at ${F}$ and ${G}$, respectively. Prove that ${D, E, F}$ and ${G}$ are collinear.
Problem
Source: Nordic Mathematical Contest 2015 #1
Tags: geometry
madmathlover
23.09.2017 18:14
Lemma: Let the incircle touches the sides $BC, CA$ of $\triangle ABC$ at $F, G$, respectively. Let $AI \cap FG = D_1$ and $BI \cap FG = E_1$. Then $\angle AD_1B = \angle AE_1B = 90^{\circ}$. Proof: well known. So, $E_1 = E$ and $F_1 = F$ and the result follows.
dikhendzab
09.04.2020 15:03
Lines $AD$ and $AC$ intersect each other at $I$, the incenter of triangle $ABC$. So, quadrilateral $AIGE$ is cyclic. Now, angles on the same arc are equal and by thales theorem $\angle AEI=\angle AEB=\angle AGI=90^{\circ}$, so $\angle IEG=\angle BEG=\angle IAG=\angle CAD=\angle BAD=\angle DEB \implies$ $G, D, E$ are collinear and $F, E, D$ are collinear, so all points $E, F, G, D$ are collinear.
primesarespecial
08.06.2021 14:54
The lemma from EGMO chapter 1 about the line through two intouch points and the angles bisector justt solves the problem.