Problem (2022IZhautykovOp3): In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.
AlphaGeometry formalization: a b c = triangle; d = parallelogram a b c d; m = on_bline b c, eqdistance m b a b; n = on_line c m, on_line a d; k = reflect n m d; l = on_line m k, on_line a d; o1 = circumcenter a m d; o2 = circumcenter c n k; p = on_circle o1 a, on_circle o2 c ? eqangle c p p m d p p l
HAGeo formalization: A B C = triangle; D = parallelogram A B C; l1 = perpendicular_bisector B C; (B) = circle_center_point B A; M = intersection l1 (B); CM = line C M; AD = line A D; N = intersection CM AD; MD = line M D; K = reflect_point_wrt_line N MD; MK = line M K; L = intersection MK AD; O1 = circumcenter A M D; O2 = circumcenter C N K; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 C; P = intersection (O1) (O2); Prove: equal_angle C P M D P L
Difficulty score: 4.0