Question

If the medians of a triangle $ABC$ intersect at $G,$ prove that:
$ar.\left(\Delta AGB\right)=ar.\left(\Delta AGC\right)=ar\left(\Delta BGC\right)=\frac{1}{3}ar\left(\Delta ABC\right)$

Answer 1

Given,

AM, BN and CL are medians.

To prove:

$ar\left(\Delta AGB\right)=ar\left(\Delta AGC\right)=ar\left(\Delta BGC\right)$

$=\frac{1}{3}.ar\left(\Delta ABC\right)$

Proof:

To $\Delta AGB$ & $\Delta AGC$

AG is the median

$\therefore$ $ar.\Delta AGB=ar.\Delta AGC$

Similarly

BG is the median

$\therefore ar.\Delta AGB=ar.\Delta BGC$

So we can say that

$ar.\Delta AGB=ar.\Delta AGC=ar.\Delta BGC$

Now,

$\Delta AGB+\Delta AGC+\Delta BGC=ar.\Delta ABC$

$\frac{1}{3}.\Delta AGB+\frac{1}{3}\Delta AGC+\frac{1}{3}\Delta BGC=ar.\Delta ABC$

(they are in equal area)

$\Rightarrow \frac{1}{3}(\Delta AGB+\Delta AGC+\Delta BGC)=ar.\Delta ABC$

$\Rightarrow \Delta AGB+\Delta AGC+\Delta BGC=\frac{1}{3}.ar\Delta ABC$

Hence proved.