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Slovenian Mathematical Olympiad, 2016

Geometry Difficulty 4.5 AIME Prove it Slovenia

Let $ABC$ be a triangle such that $|AB| = 2|AC|$ and let $D$ be a point on the ray $CA$ such that $|CD| = 3|AC|$ . Prove that the line from $C$ perpendicular to the line $BD$ bisects the segment $AB$ .

Solution

Denote the intersection of the line through $C$ perpendicular to $BD$ and the line $BD$ by $E$ and the intersection of the lines $CE$ and $AB$ by $M$ .

We have $|AD| = 2|AC| = |AB|$ , so the triangle $DBA$ is isosceles with the apex at $A$ , and so $\angle BDA = \angle ABD$ .

This implies $\angle ACM = 90^\circ - \angle EDC = 90^\circ - \angle BDA = 90^\circ - \angle ABD = \angle EMB = \angle CMA$ .

So, the triangle $CAM$ is also isosceles with the apex at $A$ . From here we get $|AM| = |AC| = \frac{1}{2}|AB|$ , which implies that $M$ is the midpoint of the segment $AB$ .

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