Gravity Center(2)

When you draw the line from A through G, the intersection of two medians from the other two vertices, the line will be the last midpoint…

To explain the above axiom, we will first explain that the intersection G of two medians BG:GN = CG:GL = 2:1.

If only the case of BG:GN is explained, the same explanation will be valid for the other.

For the discussion a supplementary line is needed. Draw it from the midpoint L to AC, parallel to BN. Let K to denote the intersection of the line and AC.

As AL=LB and LK//BN, AK=KN (This requires a proof, two, but I’d like to skip it, because it would be visibly understandable)

Then it is derived that CN=2NK=2KA, as N is the midpoint of AC.

And BN=2LK, as the triangle ALK is just 1/2 scale of the triangle ABN.

Let’s look at LK in a different triangle, CLK, including a smallaer triangle CGN.

As CN=2NK, CK=3NK.
As G is a point on BN, LK//GN. So the triangle CGN is 2/3 scale of the triangle CLK.
So, LK:GN=3:2.

As BN=2LK, BN:LK:GN=6:3:2.
As BG=BN-GN, BG:GN=4:2=2:1.

In this way, it was proved that BG:GN=2:1 .
It’s a crazy problem of geometry.


6 thoughts on “Gravity Center(2)

    • Hobby, yes. It’s my hobby to learn and share the joy of learning. When I had this problem of the gravity center, I did not see how to prove it at all. Then my teacher gave me an excellent explanation, that made me very happy.

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