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Romanian Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Romania

Let $ABCD$ be a square and consider the points $K \in (AB)$ , $L \in (BC)$ , and $M \in (CD)$ such that $KLM$ is a right isosceles triangle, with the right angle at $L$ . Prove that the lines $AL$ and $DK$ are perpendicular to each other.

Bogdan Enescu

Solution

It is not difficult to observe that $\triangle KLB \equiv \triangle LMC$ , hence $KB = LC$ . Because $AB = BC$ , it follows that $AK = BL$ . But then, $\triangle AKD \equiv \triangle BLA$ , and since $AK \perp BL$ and $AD \perp BA$ , we deduce that $AL \perp KD$ , as well.

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