In what ratio does the centroid of the triangle divide each median of the triangle?

Given G is centroid,

AD,BE,CF are median as shown in figure.

To Prove,

$\frac{\mathrm{AG}}{\mathrm{GD}}=\frac{\mathrm{BG}}{\mathrm{GE}}=\frac{\mathrm{CG}}{\mathrm{GF}}=\frac{2}{1}$

Construction : Produce AD to K such that AG=GK, join BK and CK

Proof : In △ABK,

F and G are mid points of AB and AK respectively

So, FG∥BK [by the mid point theorem]

Hence we can say that GC ∥ BK ..... (A)

In △AKC

Similarly, BG ∥ KC ..... (B)

By (A) and (B)

BGCK is a parallelogram 

In a parallelogram, diagonals bisect each other 

So, GD = DK------(C)

AG=GK  [By construction]

AG=GD+DK

So, AG=2GD [By (C)]

$\Rightarrow \frac{AG}{GD}=\frac{2}{1}$

Thus, the centroid of the triangle divides each of its median in the ratio 2:1