Problem (2009G8): Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_1$, $I_2$ and $I_3$ the incenters of $\triangle ABM$, $\triangle MNC$ and $\triangle NDA$, respectively. Prove that the orthocenter of $\triangle I_1I_2I_3$ lies on $g$.
AlphaGeometry formalization: t1 t2 t3 = triangle; i = circumcenter t1 t2 t3; t4 = on_circle i t1; a = on_tline t1 i t1, on_tline t2 i t2; b = on_tline t2 i t2, on_tline t3 i t3; c = on_tline t3 i t3, on_tline t4 i t4; d = on_tline t1 i t1, on_tline t4 i t4; m = on_line b c; n = on_line a m, on_line c d; i1 = incenter a b m; i2 = incenter m n c; i3 = incenter n d a; h = orthocenter i1 i2 i3 ? coll h a m
HAGeo formalization: T1 T2 T3 = triangle; I = circumcenter T1 T2 T3; (I) = circle_center_point I T1; T4 = on_circle (I); IT1 = line I T1; IT2 = line I T2; IT3 = line I T3; IT4 = line I T4; l1 = perpendicular_line T1 IT1; l2 = perpendicular_line T2 IT2; A = intersection l1 l2; l3 = perpendicular_line T3 IT3; B = intersection l2 l3; l4 = perpendicular_line T4 IT4; C = intersection l3 l4; D = intersection l1 l4; BC = line B C; M = on_line BC; AM = line A M; CD = line C D; N = intersection AM CD; I1 = incenter A B M; I2 = incenter M N C; I3 = incenter N D A; H = orthocenter I1 I2 I3; Prove: collinear H A M
Difficulty score: 4.0