Problem (2024ARMOg9p4): In cyclic quadrilateral $ABCD$, $\angle A+ \angle D=\frac{\pi}{2}$. $AC$ intersects $BD$ at ${E}$. A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$, prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c, on_tline c a b; e = on_line a c, on_line b d; z = on_line a c, eqdistance z a c e; t = on_line d b, eqdistance t d b e; x = on_line a b, on_line z t; y = on_line c d, on_line z t; o1 = circumcenter e z t; e1 = mirror e o1 ? cong e e1 x y
HAGeo formalization: A B C = triangle; (ABC) = circumcircle A B C; AB = line A B; l1 = perpendicular_line C AB; D = intersection (ABC) l1; AC = line A C; BD = line B D; E = intersection AC BD; (A) = circle_center_radius A C E; Z = intersection AC (A); (D) = circle_center_radius D B E; DB = line D B; T = intersection DB (D); AB = line A B; ZT = line Z T; X = intersection AB ZT; CD = line C D; Y = intersection CD ZT; O1 = circumcenter E Z T; E1 = reflect_point_wrt_point E O1; Prove: cong E E1 X Y
Difficulty score: 4.1