The final part of explanation on the gravity center of a triangle.

Assume in the below figure that AL=LB and CN=NA.

AM is drawn to join A and G.

But BM:MC is still unknown.

We are going to prove that BM=MC.

We need a supplementary line also this time.

It is drawn from the midpoint L on AB, being parallel to AM.

Let J to denote the intersection of the line with BC.

That is, LJ//AM.

So it is obvious that **BJ=JM**, because BL=LA.

The we think of another triangle including LG and JM, that is, the triangle CLJ.

In the above figure, it has already been proved that CG:GL=2:1.

Then CM:MJ = 2:1 is derived, because GM, a part of AM, is parallel to LJ.

CM=2MJ

BM=BJ+MJ

and BJ=MJ

So BM=2MJ=CM

Thus it is proved that three medians cross each other with a single point G.

You might think it a meaningless trial to explain where such and such point would be located or what length a line would have. But to prove a geometrical property under a certain condition is to prove that you could control the geometry by fixing conditions. You could get the one and only center of gravity for a given triangle, and triangles are fundamental elements of all polygons.