Problem (2024AsiaPacificMOp5): Line $\ell$ intersects sides $BC$ and $AD$ of cyclic quadrilateral $ABCD$ in its interior points $R$ and $S$, respectively, and intersects ray $DC$ beyond point $C$ at $Q$, and ray $BA$ beyond point $A$ at $P$. Circumcircles of the triangles $QCR$ and $QDS$ intersect at $N \neq Q$, while circumcircles of the triangles $PAS$ and $PBR$ intersect at $M\neq P$. Let lines $MP$ and $NQ$ meet at point $X$, lines $AB$ and $CD$ meet at point $K$ and lines $BC$ and $AD$ meet at point $L$. Prove that point $X$ lies on line $KL$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; p1 q1 = segment; p = on_line a b, on_line p1 q1; q = on_line c d, on_line p1 q1; r = on_line b c, on_line p1 q1; s = on_line a d, on_line p1 q1; o1 = circumcenter q c r; o2 = circumcenter q d s; n = on_circle o1 q, on_circle o2 q; o3 = circumcenter p a s; o4 = circumcenter p b r; m = on_circle o3 p, on_circle o4 p; x = on_line m p, on_line n q; k = on_line a b, on_line c d; l = on_line b c, on_line a d ? coll x k l
HAGeo formalization: A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); l = line; BC = line B C; R = intersection l BC; AD = line A D; S = intersection l AD; CD = line C D; Q = intersection l CD; AB = line A B; P = intersection l AB; O1 = circumcenter C Q R; (CQR) = circle_center_point O1 Q; O2 = circumcenter D Q S; (DQS) = circle_center_point O2 Q; N = intersection (CQR) (DQS); O3 = circumcenter A P S; (APS) = circle_center_point O3 P; O4 = circumcenter B P R; (BPR) = circle_center_point O4 P; M = intersection (APS) (BPR); MP = line M P; NQ = line N Q; X = intersection MP NQ; K = intersection AB CD; L = intersection BC AD; KL = line K L; Prove: collinear X K L
Difficulty score: 4.1