Problem (2011CGMOp8): The $A$-excircle $(O)$ of $\triangle ABC$ touches $BC$ at $M$. The points $D,E$ lie on the sides $AB,AC$ respectively such that $DE\parallel BC$. The incircle $(O_1)$ of $\triangle ADE$ touches $DE$ at $N$. If $BO_1\cap DO=F$ and $CO_1\cap EO=G$, prove that the midpoint of $FG$ lies on $MN$.
AlphaGeometry formalization: a b c = triangle; o = excenter a b c; m = foot o b c; d = on_line a b; e = on_line a c, on_pline d b c; o1 = incenter a d e; n = foot o1 d e; f = on_line b o1, on_line d o; g = on_line c o1, on_line e o; m1 = midpoint f g ? coll m1 m n
HAGeo formalization: A B C = triangle; O = excenter A B C; BC = line B C; M = foot O BC; AB = line A B; D = on_line AB; l1 = parallel_line D BC; AC = line A C; E = intersection AC l1; O1 = incenter A D E; DE = line D E; N = foot O1 DE; BO1 = line B O1; DO = line D O; F = intersection BO1 DO; CO1 = line C O1; EO = line E O; G = intersection CO1 EO; L = midpoint F G; Prove: collinear L M N
Difficulty score: 4.3