Problem (2022IranGOMp3): Let $O$ be the circumcenter of triangle $ABC$. Arbitrary points $M$ and $N$ lie on the sides $AC$ and $BC$, respectively. Points $P$ and $Q$ lie in the same half-plane as point $C$ with respect to the line $MN$, and satisfy $\triangle CMN \sim \triangle PAN \sim \triangle QMB$ (in this exact order). Prove that $OP=OQ$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; m = on_line a c; n = on_line b c; p = on_aline p a n c m n, on_aline p n a c n m; q = on_aline q m b c m n, on_aline q b m c n m ? cong o p o q
HAGeo formalization: A B C = triangle; O = circumcenter A B C; AC = line A C; M = on_line AC; BC = line B C; N = on_line BC; l1 = angle_equal1 A N C M N; l2 = angle_equal1 N A C N M; P = intersection l1 l2; l3 = angle_equal1 M B C M N; l4 = angle_equal1 B M C N M; Q = intersection l3 l4; Prove: cong O P O Q
Difficulty score: 3.2