Problem (2022G2): In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$. Prove that $B, C, X,$ and $Y$ are concyclic.
AlphaGeometry formalization: a b c = triangle; p = on_tline a b c; d = on_line b c, on_pline p a c; e = on_line b c, on_pline p a b; o1 = circumcenter a b d; o2 = circumcenter a c e; x = on_circle d a, on_circle o1 a; y = on_circle e a, on_circle o2 a ? cyclic b c x y
HAGeo formalization: A B C = acute_triangle; BC = line B C; l = perpendicular_line A BC; P = on_line l; AC = line A C; l1 = parallel_line P AC; D = intersection l1 BC; AB = line A B; l2 = parallel_line P AB; E = intersection l2 BC; O1 = circumcenter A B D; (O1) = circle_center_point O1 A; (D) = circle_center_point D A; X = intersection (O1) (D); O2 = circumcenter A C E; (O2) = circle_center_point O2 A; (E) = circle_center_point E A; Y = intersection (O2) (E); Prove: concyclic B C X Y
Difficulty score: 2.4