<Tri-2025-2235>, hint

Suppose a line through O(=G), parallel to BC, meet AB, QM, PN, AC 

at B', Q', P', C'  respectively. AG(=AO) meets BC at A', BA'=A'C.

AO/OA'=2/1, OM=OP, hence AP=1/3 AC.

B'Q'+Q'O=OP'+P'C (BC//B'C'), since Q'O=P'O, B'Q'=P'C'. Hence BA'=CA', BM=CN

ABC is isosceles.