Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABC$ be a triangle and $P$ a point inside it. Rays $BP$ and $CP$ meet $AC$ and $AB$ at $Y$ and $X$ , respectively. Prove that if $AP$ bisects $BC$ then $XY \parallel BC$ .

Solution

Solution:

Let $Q$ be the reflection of $P$ across $M$ (with $M$ the midpoint of $BC$ ). Accordingly, $BPCQ$ is a parallelogram.

From this, we see that $\triangle AXP \sim \triangle ABQ$ and $\triangle AYP \sim \triangle ACQ$ , and thus we deduce

$\frac{AX}{AB} = \frac{AP}{AQ} = \frac{AY}{AC}$

so $XY \parallel BC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.