A point $ N $ is marked on the median $ CM $ of the triangle $ ABC $ so that $ MN \cdot MC = AB ^ 2/4 $. Lines $ AN $ and $ BN $ intersect the circumcircle $ \triangle ABC $ for the second time at points $ P $ and $ Q $, respectively. $ R $ is the point of segment $ PQ $, nearest to $ Q $, such that $ \angle NRC = \angle BNC $. $ S $ is the point of the segment $ PQ $ closest to $ P $ such that $ \angle NSC = \angle ANC $. Prove that $ RN = SN $.