Problem (2008CTSTp13): Let $ ABC$ be an acute triangle, let $ M,N$ be the midpoints of minor arcs $ \widehat{CA},\widehat{AB}$ of the circumcircle of triangle $ ABC,$ point $ D$ is the midpoint of segment $ MN,$ point $ G$ lies on minor arc $ \widehat{BC}.$ Denote by $ I,I_{1},I_{2}$ the incenters of triangle $ ABC,ABG,ACG$ respectively.Let $ P$ be the second intersection of the circumcircle of triangle $ GI_{1}I_{2}$ with the circumcircle of triangle $ ABC.$ Prove that three points $ D,I,P$ are collinear.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; m = on_circum a b c, on_bline a c; n = on_circum a b c, on_bline a b; d = midpoint m n; g = on_circum a b c; i = incenter a b c; i1 = incenter a b g; i2 = incenter a c g; o1 = circumcenter g i1 i2; p = on_circum g i1 i2, on_circum a b c ? coll d i p
HAGeo formalization: A B C = acute_triangle; O = circumcenter A B C; M = midarc_no C A B; N = midarc_no A B C; D = midpoint M N; (ABC) = circle A B C; G = on_circle (ABC); I = incenter A B C; I1 = incenter A B G; I2 = incenter A C G; O1 = circumcenter G I1 I2; (GI1I2) = circle G I1 I2; P = intersection (GI1I2) (ABC); Prove: collinear D I P
Difficulty score: 3.1