Track / Stage 6 / 30 of 400 #1030 of 2000

Problem 1030

National olympiad, first round

Geometry Difficulty 6.0 Prove it

## Problem 1.

Let $A B C$ be a triangle and $\Gamma$ the circle with diameter $A B$ . The bisectors of $\angle B A C$ and $\angle A B C$ intersect $\Gamma$ (also) at $D$ and $E$ , respectively. The incircle of $A B C$ meets $B C$ and $A C$ at $F$ and $G$ , respectively. Prove that $D, E, F$ and $G$ are collinear.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution 1. Let the line $E D$ meet $A C$ at $G^{\prime}$ and $B C$ at $F^{\prime} . A D$ and $B E$ intersect at $I$ , the incenter of $A B C$ . As angles subtending the same arc $\widehat{B D}$ , $\angle D A B=\angle D E B=\angle G^{\prime} E I$ . But $\angle D A B=\angle C A D=$ $\angle G^{\prime} A I$ . This means that $E, A, I$ and $G^{\prime}$ are concyclic, and $\angle A E I=\angle A G^{\prime} I$ as angles subtending the same chord $A I$ . But $A B$ is a diameter of $\Gamma$ , and so $\angle A E B=$ $\angle A E I$ is a right angle. So $I G^{\prime} \perp A C$ , or $G^{\prime}$ is the foot of the perpendicular from $I$ to $A C$ . This implies $G^{\prime}=G$ . In a similar manner we prove that $F^{\prime}=F$ , and the proof is complete.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.