Note that $MA^2 = MN \cdot MC = MB^2$, so $\triangle MAC \sim \triangle MNA$ and $\triangle MBC \sim \triangle MNB$. We see that $\angle CPN = \angle CPA = \angle CBA = \angle CBM = \angle BNM = 180^{\circ} - \angle BNC = 180^{\circ} - \angle CRN$. Thus, $CPNR$ is cyclic. Similarly, we can show $CQNS$ is cyclic. Also, we see that $\angle CNP = \angle ANM = \angle CAM = \angle CAB$ and $\angle CPN = \angle CPA = \angle CBA$, so $\triangle CNP \sim \triangle CAB$. Hence, $\angle NCP = \angle ACB$. Similarly, we can show that $\triangle CQN \sim \triangle CAB$ and $\angle QCN = \angle ACB$. Thus, $\angle NRS = \angle NRP = \angle NCP = \angle ACB = \angle QCN = \angle QSN = \angle RSN$, so $RN = SN$.