Question

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

Answer 1

Let $ABC$ be any triangle whose vertices $A,B$ and $C$ respectively.

We consider that $D,E$ and $F$ be mid point of $\Delta ABC$ and $G$ be the centroid $\Delta ABC$

Then, $AE,BF\&CD$ are medians

To prove- area $\Delta AGB=$ area $\Delta BGC$ area $\Delta AGB=$ area $\Delta BGC=$ area $\Delta CGA=\frac{1}{3}$ area $\Delta ABC$

Now,

In $\Delta AGB$ and $\Delta AGC$

$AG$ is the median

Therefore ,

$ar\Delta AGB=ar\Delta AGC$ …….. (1)

Similarly,

$BG$ is the median

Therefore,

$ar\Delta AGB=ar\Delta BGC$ …… (2)

By equation $\left(1\right)\&\left(2\right)$ to,

$ar\Delta AGB=ar\Delta AGC=ar\Delta BGC$ ……. (3)

Now, We know that

$ar\Delta AGB+ar\Delta BGC+ar\Delta CGA=ar\Delta ABC$ ......(4)

By equation $\left(3\right)$ and $\left(4\right)$ to, we get

$ar\Delta AGB+ar\Delta AGB+ar\Delta AGB=ar\Delta ABC,$

$ar\Delta AGB=\frac{1}{3}ar△ABC$

Substituting in equation(1),

$ar\Delta AGB=ar\Delta BGC=ar\Delta CGA=\frac{1}{3}ar\Delta ABC$

Hence proved.