Problem (2023CHNSouthEastMOg10p3): In acute triangle $ABC$ ($\triangle ABC$ is not an isosceles triangle), $I$ is its incentre, and circle $ \omega$ is its inscribed circle. $\odot\omega$ touches $BC, CA, AB$ at $D, E, F$ respectively. $AD$ intersects with $\odot\omega$ at $J$ ($J\neq D$), and the circumcircle of $\triangle BCJ$ intersects $\odot\omega$ at $K$ ($K\neq J$). The circumcircle of $\triangle BFK$ and $\triangle CEK$ meet at $L$ ($L\neq K$). Let $M$ be the midpoint of the major arc $BAC$.Prove that $M, I, L$ are collinear.
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; j = on_line a d, on_circle i d; o1 = circumcenter b c j; k = on_circle o1 b, on_circle i d; o2 = circumcenter b f k; o3 = circumcenter c e k; l = on_circle o2 b, on_circle o3 c; o = circumcenter a b c; m = on_bline b c, on_circle o a ? coll m i l
HAGeo formalization: A B C = acute_triangle; 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; J = intersection AD (I); O1 = circumcenter B C J; (BCJ) = circle_center_point O1 B; K = intersection (BCJ) (I); O2 = circumcenter B F K; (BFK) = circle_center_point O2 B; O3 = circumcenter C E K; (CEK) = circle_center_point O3 C; L = intersection (BFK) (CEK); M = midarc B A C; Prove: collinear M I L
Difficulty score: 5.3