Problem (2021G4): Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; o = circumcenter a b c; e = on_line b a, on_tline d o d; f = on_line b c, on_tline d o d; t = on_pline e c d, on_pline f a d; k = on_line d f, on_circle t d; x = on_line a c, on_line d t ? coll b k x
HAGeo formalization: A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); O = circumcenter A B C; OD = line O D; l = perpendicular_line D OD; AB = line A B; E = intersection l AB; BC = line B C; F = intersection l BC; CD = line C D; l1 = parallel_line E CD; AD = line A D; l2 = parallel_line F AD; T = intersection l1 l2; DF = line D F; (T,D) = circle_center_point T D; K = intersection DF (T,D); AC = line A C; DT = line D T; X = intersection AC DT; Prove: collinear B K X
Difficulty score: 3.5