Problem (2023G2): Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; p = on_line a c, on_circle c b; s = foot p a b; d = on_line b p, on_circle o a; q = on_line s p, eqdistance p o a; e = on_tline a c q, on_tline b d q ? cyclic e a b c
HAGeo formalization: A B C = triangle; O = circumcenter A B C; AC = line A C; (C) = circle_center_point C B; P = intersection AC (C); AB = line A B; S = foot P AB; BP = line B P; (O) = circle_center_point O A; D = intersection BP (O); PS = line P S; (P,AO) = circle_center_radius P A O; Q = intersection PS (P,AO); CQ = line C Q; l1 = perpendicular_line A CQ; DQ = line D Q; l2 = perpendicular_line B DQ; E = intersection l1 l2; Prove: concyclic E A B C
Difficulty score: 2.3