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Geometry Difficulty 5.0 AIME, harder Prove it Turkey

Let $ABCD$ be a cyclic quadrilateral and let the midpoints of $AB$ , $BC$ , $CD$ and $DA$ be $K$ , $L$ , $M$ and $N$ , respectively. Let the reflections of the point $M$ with respect to the lines $AD$ and $BC$ be $P$ and $Q$ , respectively. Finally the circumcenter of the triangle $KPQ$ be $R$ . Prove that $RN = RL$ .

Solution

We start by noting that $KLMN$ is a parallelogram, and by the symmetry we have $NP = NM = KL$ and $LQ = LM = KN$ . Moreover, we have $\angle KNM = \angle KLM$ .

Since $ABCD$ is cyclic, $MN \parallel AC$ and $LM \parallel BD$ we get that $\angle DNM = \angle DAC = \angle DBC = \angle MLC$ and from the symmetry we get $\angle MNP = 2\angle MND = 2\angle MLC = \angle MLQ$ and finally we obtain $\angle KNP = \angle KLQ$ .

Combining this with the side equalities above we find $\angle KNP \cong \angle QLK$ and hence $KP = KQ$ . Since $R$ is the circumcenter of the triangle $KPQ$ we also have $\angle RPK \cong \angle RKQ$ and hence $\angle RKN \cong \angle RQL$ which implies $RN = RL$ . We are done.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.