Team Selection Test for JBMO 2024

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.