Problem (2016G2): Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; i = incenter a b c; m = midpoint b c; d = foot i b c; e = on_line a c, on_tline i a i; f = on_line a b, on_tline i a i; o1 = circumcenter a e f; x = on_circle o1 a, on_circle o a; y = on_line x d, on_line a m ? cyclic y a b c
HAGeo formalization: A B C = triangle; I = incenter A B C; M = midpoint B C; BC = line B C; D = foot I BC; AI = line A I; l = perpendicular_line I AI; AB = line A B; F = intersection l AB; AC = line A C; E = intersection l AC; O1 = circumcenter A E F; (O1) = circle_center_point O1 A; O = circumcenter A B C; (O) = circle_center_point O A; X = intersection (O1) (O); DX = line D X; AM = line A M; T = intersection DX AM; Prove: concyclic A B C T
Difficulty score: 3.0