Problem (2023G5): Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$. Let $CO$ meet $AB$ at $X$, and $BO$ meet $AC$ at $Y$. Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; d = on_circle o a, on_tline b a c; e = on_circle o a, on_tline c a b; x = on_line c o, on_line a b; y = on_line b o, on_line a c; o1 = circumcenter b x d; o2 = circumcenter c y e; p = on_circle o1 b, on_circle o2 c ? coll p a o
HAGeo formalization: A B C = acute_triangle; O = circumcenter A B C; (O) = circle_center_point O A; AC = line A C; l1 = perpendicular_line B AC; D = intersection (O) l1; AB = line A B; l2 = perpendicular_line C AB; E = intersection (O) l2; CO = line C O; X = intersection CO AB; BO = line B O; Y = intersection BO AC; O1 = circumcenter B D X; O2 = circumcenter C E Y; (O1) = circle_center_point O1 B; (O2) = circle_center_point O2 C; Z = intersection (O1) (O2); Prove: collinear A O Z
Difficulty score: 3.9