Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Iran

Let $ABCD$ be a rhombus and let $\omega$ be its incircle. Let $M$ be the midpoint of $AB$ and $K$ be a point inside $ABCD$ such that $MK$ is tangent to $\omega$ . Prove that $CDKM$ is cyclic.

Solution

Let $L$ be the intersection of $KM$ and $CD$ . The quadrilateral $MLCB$ has an incircle centered at $O$ . So $OL$ and $OM$ are angle-bisectors of $\angle MLC$ and $\angle BML$ , respectively. Further, $BM \parallel CL$ , these two arguments yielding $\angle LOM = 90^\circ$ . Therefore $LO$ is tangent to the circumcircle of $OKM$ , since $\angle MKO = 90^\circ$ . Moreover, $\angle LOM = \angle AOD$ implies that
$\angle LOD = \angle AOM = \angle BAC = \angle LCO.$

Hence $LO$ is also tangent to the circumcircle of the triangle $DOC$ . Therefore we would have
$LD \cdot LC = LO^2 = LK \cdot LM,$
so $MKDC$ is cyclic. As desired.

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