Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it

1. In a non-isosceles $\triangle ABC$ , the incenter is $I$ , and the incircle $\odot I$ touches $BC$ , $CA$ , and $AB$ at points $A_{1}$ , $B_{1}$ , and $C_{1}$ , respectively. $AA_{1}$ and $BB_{1}$ intersect $\odot I$ again at points $A_{2}$ and $B_{2}$ . In $\triangle A_{1}B_{1}C_{1}$ , the angle bisectors of $\angle A_{1}$ and $\angle B_{1}$ intersect $B_{1}C_{1}$ and $C_{1}A_{1}$ at points $A_{3}$ and $B_{3}$ , respectively. Prove that $A_{2}A_{3}$ and $B_{2}B_{3}$ are the angle bisectors of $\angle B_{1}A_{2}C_{1}$ and $\angle C_{1}B_{2}A_{1}$ , respectively.

Solution

From property 2, we know that quadrilateral $A_{2} C_{1} A_{1} B_{1}$ is a harmonic quadrilateral. Therefore,
$\frac{A_{2} C_{1}}{B_{1} A_{2}}=\frac{A_{1} C_{1}}{A_{1} B_{1}}=\frac{A_{3} C_{1}}{A_{3} B_{1}} .$

Thus, $A_{2} A_{3}$ is the angle bisector of $\angle B_{1} A_{2} C_{1}$ .
Similarly, $B_{2} B_{3}$ is the angle bisector of $\angle C_{1} B_{2} A_{1}$ .

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