HMMT February, 2024

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
Let $ABC$ be a scalene triangle and $M$ be the midpoint of $BC$ . Let $X$ be the point such that $CX \parallel AB$ and $\angle AMX = 90^{\circ}$ . Prove that $AM$ bisects $\angle BAX$ .

Solution

Solution:

Let $Y$ be the intersection of lines $AB$ and $XM$ . Since $BY \parallel CX$ , we have $\angle YBM = \angle XCM$ . Furthermore, we have $BM = CM$ , since $M$ is the midpoint of $BC$ . Thus,
$\triangle BMY \cong \triangle CMX$
Thus, $MY = MX$ . Combined with the condition $AM \perp XY$ , we get that $AYX$ is an isosceles triangle with median $AM$ . Therefore, $AM$ bisects $\angle YAX$ which is the same as $\angle BAX$ and we are done.

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