Problem (2010CTSTp4): Let $\triangle ABC$ be an acute triangle with $AB>AC$, let $I$ be the center of the incircle. Let $M,N$ be the midpoint of $AC$ and $AB$ respectively. $D,E$ are on $AC$ and $AB$ respectively such that $BD\parallel IM$ and $CE\parallel IN$. A line through $I$ parallel to $DE$ intersects $BC$ in $P$. Let $Q$ be the projection of $P$ on line $AI$. Prove that $Q$ is on the circumcircle of $\triangle ABC$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; m = midpoint a c; n = midpoint a b; d = on_line a c, on_pline b i m; e = on_line a b, on_pline c i n; p = on_line b c, on_pline i d e; q = foot p a i ? cyclic q a b c
HAGeo formalization: A B C = acute_triangle; I = incenter A B C; M = midpoint A C; N = midpoint A B; IM = line I M; l1 = parallel_line B IM; AC = line A C; D = intersection AC l1; IN = line I N; l2 = parallel_line C IN; AB = line A B; E = intersection AB l2; DE = line D E; DE = parallel_line I DE; l3 = parallel_line I DE; BC = line B C; P = intersection l3 BC; AI = line A I; Q = foot P AI; Prove: concyclic Q A B C
Difficulty score: 4.2