Problem (2020CSMGOp6): Let $\triangle ABC$ be a triangle with the incenter $I$, and let the incircle $\omega$ of $\triangle ABC$ touch $\overline{BC},\overline{CA},\overline{AB}$ at points $D,E,F$ respectively. Let $\overline{AD}$ intersect $\omega$ again at a point $X\ne D$. Let the lines through $E$ and $F$ parallel to $AD$ intersect $\omega$ again at points $P\ne E, Q\ne F$ respectively. Prove that $I$ lies on the common chord of the circumcircle of $\triangle XBQ$ and the circumcircle of $\triangle XCP$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; x = on_line a d, on_circle i d; p = on_pline e a d, on_circle i d; q = on_pline f a d, on_circle i d; o1 = circumcenter x b q; o2 = circumcenter x c p; k = on_circle o1 x, on_circle o2 x ? coll i x k
HAGeo formalization: A B C = triangle; I = incenter A B C; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; AD = line A D; X = intersection AD (I); l1 = parallel_line E AD; P = intersection l1 (I); l2 = parallel_line F AD; Q = intersection l2 (I); O1 = circumcenter B Q X; (BQX) = circle_center_point O1 X; O2 = circumcenter C P X; (CPX) = circle_center_point O2 X; U = intersection (BQX) (CPX); Prove: collinear I U X
Difficulty score: 5.1