Problem (2024EGMOp2): Let $ABC$ be a triangle with $AC>AB$ , and denote its circumcircle by $\Omega$ and incentre by $I$. Let its incircle meet sides $BC,CA,AB$ at $D,E,F$ respectively. Let $X$ and $Y$ be two points on minor arcs $\widehat{DF}$ and $\widehat{DE}$ of the incircle, respectively, such that $\angle BXD = \angle DYC$. Let line $XY$ meet line $BC$ at $K$. Let $T$ be the point on $\Omega$ such that $KT$ is tangent to $\Omega$ and $T$ is on the same side of line $BC$ as $A$. Prove that lines $TD$ and $AI$ meet on $\Omega$.
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; o = circumcenter a b c; x = on_circle i d; y = on_circle i d, eqangle3 y d c x b d; k = on_line x y, on_line b c; t t1 = tangent k o a; u = on_line t d, on_line a i ? cyclic a b c u
HAGeo formalization: A B C = triangle; I = incenter A B C; BC = line B C; CA = line C A; AB = line A B; D = foot I BC; E = foot I CA; F = foot I AB; (I) = circle_center_point I D; X = on_circle (I); ω1 = angle_equal2 D C B X D; Y = intersection ω1 (I); XY = line X Y; K = intersection XY BC; O = circumcenter A B C; (O) = circle_center_point O A; T T1 = tangent_point K (O); TD = line T D; AI = line A I; U = intersection TD AI; Prove: concyclic A B C U
Difficulty score: 2.2