Hehe, there was a mistake just now. It should be that the first one is the center of gravity (the intersection of the center lines) and the fourth one is the outer center (the center of the circumscribed circle).

First, according to the known OA+OB+OC=0,

That is to say, OA+OB=-OC, and OA+OB is represented by parallelogram rule, that is, OA OB is known.

The diagonal of the parallelogram is on the extension line of CO, which bisects the AB side (because the diagonal of the parallelogram bisects each other, AB is just the other diagonal of the parallelogram), that is, CO is the center line of the AB side, similarly BO is the center line of the AC side, and AO is the center line of the BC side.

Fourthly, it is known that (OA+OB)AB=0, the vector (OA+OB) is perpendicular to AB, and OA+OB is represented by the parallelogram rule, that is, it is known that the diagonal of the parallelogram composed of OA+OB is perpendicular to the other diagonal AB of this parallelogram, which means that this parallelogram is a diamond, so |OA|=|OB|, and then it can be known that O is the middle of AB side by being perpendicular to AB.