If G be the centroid of a triangle ABC, prove that:

Let $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ be the coordinates of the vertices of $\triangle \mathrm{ABC}$.

Let us assume that centroid of the $\triangle \mathrm{ABC}$ is at the origin $\mathrm{G}$.

So, the coordinates of $\mathrm{G}$ are $\mathrm{G}(0,0)$.

Now, $\frac{x_{1}+x_{2}+x_{3}}{3}=0 ; \frac{y_{1}+y_{2}+y_{3}}{3}=0$

so, $x_{1}+x_{2}+x_{3}=0$.......(1)

$y_{1}+y_{2}+y_{3}=0$...........(2)

Squaring (1) and (2), we get

$x_{1}{ }^{2}+x_{2}{ }^{2}+x_{3}{ }^{2}+2 x_{1} x_{2}+2 x_{2} x_{3}+2 x_{3} x_{1}=0 \quad \ldots \ldots$ (3)

$y_{1}{ }^{2}+y_{2}{ }^{2}+y_{3}{ }^{2}+2 y_{1} y_{2}+2 y_{2} y_{3}+2 y_{3} y_{1}=0 \quad \ldots \ldots$ (4)

$\mathrm{LHS}=\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}$

$=\left[\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]^{2}+\left[\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}\right]^{2}+\left[\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}}\right]^{2}$

$=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}$

$=x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}+y_{1}^{2}+y_{2}^{2}-2 y_{1} y_{2}+x_{2}^{2}+x_{3}^{2}-2 x_{2} x_{3}+y_{2}^{2}+y_{3}^{2}-2 y_{2} y_{3}+x_{1}^{2}+x_{3}^{2}-2 x_{1} x_{3}+y_{1}^{2}+y_{3}^{2}-2 y_{1} y_{3}$

$=2\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)-\left(2 x_{1} x_{2}+2 x_{2} x_{3}+2 x_{3} x_{1}\right)-\left(2 y_{1} y_{2}+2 y_{2} y_{3}+2 y_{3} y_{1}\right)$

$=2\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)+\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)$

$=3\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)$

$\mathrm{RHS}=3\left(\mathrm{GA}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}\right)$

$=3\left[\left\{\sqrt{\left(x_{1}-0\right)^{2}+\left(y_{1}-0\right)^{2}}\right\}^{2}+\left\{\sqrt{\left(x_{2}-0\right)^{2}+\left(y_{2}-0\right)^{2}}\right\}^{2}+\left\{\sqrt{\left(x_{3}-0\right)^{2}+\left(y_{3}-0\right)^{2}}\right\}^{2}\right]$

$=3\left(x_{1}{ }^{2}+x_{2}{ }^{2}+x_{3}{ }^{2}+y_{1}{ }^{2}+y_{2}{ }^{2}+y_{3}{ }^{2}\right)$

Hence, $\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}=3\left(\mathrm{GA}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}\right)$