Problem (2024USATSTSTp4): Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ and $E$ be the intersection of segments $AC$ and $BD$. Let $\omega_1$ be the circumcircle of $ADE$ and $\omega_2$ be the circumcircle of $BCE$. The tangent to $\omega_1$ at $A$ and the tangent to $\omega_2$ at $C$ meet at $P$. The tangent to $\omega_1$ at $D$ and the tangent to $\omega_2$ at $B$ meet at $Q$. Show that $OP=OQ$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; o = circumcenter a b c; e = on_line a c, on_line b d; o1 = circumcenter a d e; o2 = circumcenter b c e; p = on_tline a o1 a, on_tline c o2 c; q = on_tline d o1 d, on_tline b o2 b ? cong o p o q
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; E = intersection AC BD; O1 = circumcenter A D E; O1A = line O1 A; l1 = perpendicular_line A O1A; O2 = circumcenter B C E; O2C = line O2 C; l2 = perpendicular_line C O2C; P = intersection l1 l2; O1D = line O1 D; l3 = perpendicular_line D O1D; O2B = line O2 B; l4 = perpendicular_line B O2B; Q = intersection l3 l4; Prove: cong O P O Q
Difficulty score: 2.7