If G be the centroid of a triangle ABC and P be any other point in the plane,

Solution:

Let $\triangle \mathrm{ABC}$ be any triangle whose coordinates are $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$. Let $\mathrm{P}$ be the origin and $\mathrm{G}$ be the centroid of the triangle.

We have to prove that,

$\mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}=\mathrm{GA}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}+3 \mathrm{GP}^{2}$

We know that the co-ordinates of the centroid $\mathrm{G}$ of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is-

$G\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{2}}{3}\right)$

In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,

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

So,

$\mathrm{PA}^{2}=\left(x_{1}-0\right)^{2}+\left(y_{1}-0\right)^{2}$

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

$\mathrm{PB}^{2}=\left(x_{2}-0\right)^{2}+\left(y_{2}-0\right)^{2}$

$=x_{2}^{2}+y_{2}^{2}$

$\mathrm{PC}^{2}=\left(x_{3}-0\right)^{2}+\left(y_{3}-0\right)^{2}$

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

Now,

$\mathrm{GP}^{2}=\left(\frac{x_{1}+x_{2}+x_{3}}{3}-0\right)^{2}+\left(\frac{y_{1}+y_{2}+y_{3}}{3}-0\right)^{2}$

$=\frac{\left(x_{1}+x_{2}+x_{3}\right)^{2}}{9}+\frac{\left(y_{1}+y_{2}+y_{3}\right)^{2}}{9}$

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

$=\frac{\left(2 x_{1}-x_{2}-x_{3}\right)^{2}}{9}+\frac{\left(2 y_{1}-y_{2}-y_{3}\right)^{2}}{9}$

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

$=\frac{\left(2 x_{2}-x_{1}-x_{3}\right)^{2}}{9}+\frac{\left(2 y_{2}-y_{1}-y_{3}\right)^{2}}{9}$

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

$=\frac{\left(2 x_{3}-x_{1}-x_{2}\right)^{2}}{9}+\frac{\left(2 y_{3}-y_{1}-y_{2}\right)^{2}}{9}$

So we get the value of left hand side of equation (1) as,

$\mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}$

Similarly we get the value of right hand side of equation (1) as,

$\mathrm{GA}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}+3 \mathrm{GP}^{2}=\left[\frac{\left(2 x_{1}-x_{2}-x_{2}\right)^{2}}{9}+\frac{\left(2 y_{1}-y_{2}-y_{3}\right)^{2}}{9}\right]+\left[\frac{\left(2 x_{2}-x_{1}-x_{3}\right)^{2}}{9}+\frac{\left(2 y_{2}-y_{1}-y_{3}\right)^{2}}{9}\right]$

$+\left[\frac{\left(2 x_{3}-x_{1}-x_{2}\right)^{2}}{9}+\frac{\left(2 y_{3}-y_{1}-y_{2}\right)^{2}}{9}\right]+3\left[\frac{\left(x_{1}+x_{2}+x_{2}\right)^{2}}{9}+\frac{\left(y_{1}+y_{2}+y_{3}\right)^{2}}{9}\right]$

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

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

Hence,

$\mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}=\mathrm{GA}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}+3 \mathrm{GP}^{2}$