Question

If $G$ is the centroid of any $ABC$, then show that $\frac{A{G}^2+B{G}^2+C{G}^2}{A{B}^2+B{C}^2+C{A}^2}=\frac{1}{3}$.

Answer 1

Let us suppose the required triangle $ABC$ is an equilateral having side length ${}^{\prime }{a}^{\prime }$.

And $G$ be centroid.

Thus $AB=BC=CA=a$

$\Rightarrow AG=BG=CG=\frac{2}{3}\times$ median length$=\frac{2}{3}\times \frac{\sqrt{3}}{2}\times a=\frac{a}{\sqrt{3}}$

$\Rightarrow \frac{A{G}^2+B{G}^2+C{G}^2}{A{B}^2+B{C}^2+C{A}^2}=\frac{3A{G}^2}{3A{B}^2}=\frac{{\left(\frac{a}{\sqrt{3}}\right)}^2}{a^2}=\frac{1}{3}$