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Slovenia — National Math Olympiad, 2012

Geometry Difficulty 4.4 AIME Prove it Slovenia

Let $ABC$ be a right-angled triangle with the right angle at $C$ . Let the points $K$ , $L$ and $M$ lie on the sides $CA$ , $AB$ and $BC$ , respectively, so that the angle $\angle MLK$ is a right angle and $|KC| = |KL|$ . Prove that the bisectors of the angles $\angle AKL$ and $\angle LMB$ are parallel.

Solution

Let $K'$ and $L'$ be the intersection points of the side $AB$ with the bisectors of the angles $\angle AKL$ and $\angle LMB$ , respectively. The triangles $KMC$ and $KML$ coincide in two sides and the angle that lies opposite the longer among the two sides, hence the triangles are congruent.

The line $KM$ is thus the bisector of the angles $\angle LKC$ and $\angle CML$ . From this we conclude that the angles $\angle K'KM$ and $\angle KML'$ are right angles. Now the claim is straightforward.

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