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Problem 837

AMC 12 late, AIME early

Geometry Difficulty 4.5 Prove it All-Soviet-Union Mathematical Olympiad · Soviet Union

$ABCD$ is a rectangle. Points $K$ , $L$ , $M$ , $N$ are chosen on $AB$ , $BC$ , $CD$ , $DA$ respectively so that $KL$ is parallel to $MN$ , and $KM$ is perpendicular to $LN$ . Show that the intersection of $KM$ and $LN$ lies on $BD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Let $LN$ and $KM$ meet at $O$ . $\angle NOM = \angle NDM = 90^{\circ}$ , so $OMDN$ is cyclic. Hence $\angle NOD = \angle NMD$ . Similarly, $BLOK$ is cyclic and $\angle LOB = \angle LKB$ . But $NM$ is parallel to $LK$ and $AB$ is parallel to $CD$ , so $\angle LKB = \angle NMD$ . Hence $\angle NOD = \angle LOB$ , so $DOB$ is a straight line.

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