Question

$ABC$ is a triangle and '$O$' is the point of intersection of the medians, then $(A{B}^2+B{C}^2+C{A}^2)$ is equal to:

Answer 1

By appolonius theorem, we have
$A{B}^2+A{C}^2=2(A{D}^2+B{D}^2)$
$\Rightarrow A{B}^2+a{c}^2=2({\left(\frac{3}{2}OA\right)}^2+{\left(\frac{BC}{2}\right)}^2)$
$\Rightarrow A{B}^2+A{C}^2=\frac{9}{2}O{A}^2+\frac{1}{2}B{C}^2$ ...(1)
Similary $B{C}^2+B{A}^2=\frac{9}{2}O{C}^2+\frac{1}{2}A{B}^2$ ...(2)
and $C{A}^2+C{B}^2=\frac{9}{2}O{C}^2+\frac{1}{2}A{B}^2$ ...(3)
Adding (1), (2) and (3), we get
$2(A{B}^2+B{C}^2+C{A}^2)=\frac{9}{2}(O{A}^2+O{B}^2+O{C}^2)+\frac{1}{2}(A{B}^2+B{C}^{2}C{A}^2)$
$\Rightarrow \frac{3}{2}(A{B}^2+B{C}^{2}C{A}^2)=\frac{9}{2}(O{A}^2+O{B}^2+O{C}^2)$
$\Rightarrow A{B}^2+B{C}^2+C{A}^2=3(O{A}^2+O{B}^2+O{C}^2)$