Problem (2020CHNSouthEastMOg11p2): In a scalene triangle $\Delta ABC$, $AB<AC$, $PB$ and $PC$ are tangents of the circumcircle $(O)$ of $\Delta ABC$. A point $R$ lies on the arc $\widehat{AC}$(not containing $B$), $PR$ intersects $(O)$ again at $Q$. Suppose $I$ is the incenter of $\Delta ABC$, $ID \perp BC$ at $D$, $QD$ intersects $(O)$ again at $G$. A line passing through $I$ and perpendicular to $AI$ intersects $AG,AC$ at $M,N$, respectively. $S$ is the midpoint of arc $\widehat{AR}$, and$SN$ intersects $(O)$ again at $T$. Prove that, if $AR \parallel BC$, then $M,B,T$ are collinear.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; p = on_tline b o b, on_tline c o c; r = on_circle o a, on_pline a b c; q = on_line p r, on_circle o a; i = incenter a b c; d = foot i b c; g = on_line q d, on_circle o a; m = on_line a g, on_tline i a i; n = on_line a c, on_tline i a i; s = on_bline a r, on_circle o a; t = on_line s n, on_circle o a ? coll m b t
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; OB = line O B; OC = line O C; l1 = perpendicular_line B OB; l2 = perpendicular_line C OC; P = intersection l1 l2; BC = line B C; l3 = parallel_line A BC; R = intersection (O) l3; PR = line P R; Q = intersection PR (O); I = incenter A B C; D = foot I BC; DQ = line D Q; G = intersection DQ (O); AI = line A I; l4 = perpendicular_line I AI; AG = line A G; M = intersection l4 AG; AC = line A C; N = intersection l4 AC; l5 = perpendicular_bisector A R; S = intersection l5 (O); NS = line N S; T = intersection NS (O); Prove: collinear M B T
Difficulty score: 3.5