If G be the centroid of a triangle ABC and O be any other point, prove that
$3 (G A^{2} + G B^{2} + G C^{2}) = B C^{2} + C A^{2} + A B^{2}$
and $O A^{2} + O B^{2} + O C^{2} = G A^{2} + G B^{2} + G C^{2}$

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Solution

Let G be the centroid of a triangle ABC.
Let D,E,F are midpoints of sides BC,CA and AB respectively.
By archimedian principle
$A B^{2} + A C^{2} = 2 [{(\frac{1}{2} B C)}^{2} + A D^{2}]$

$\Rightarrow A B^{2} + C A^{2} = 2 [\frac{B C^{2}}{4} + {(\frac{3}{2} A G)}^{2}] [∵ G i s c e n t r o i d; A G : G D = 2 : 1; \Rightarrow A G = \frac{2}{3} A D]$

$\Rightarrow A B^{2} + C A^{2} = \frac{B C^{2}}{2} + \frac{9}{2} A G^{2} ⟶ 1$

$∥$ ly $B C^{2} + B A^{2} = 2 [{(\frac{1}{2} A C)}^{2} + B E^{2}]$

$\Rightarrow B C^{2} + B A^{2} = \frac{A C^{2}}{2} + \frac{9}{2} B G^{2} ⟶ 2 [∵ B G : G E = 2 : 1]$

$∥$ ly $C A^{2} + B C^{2} = 2 [{(\frac{1}{2} A B)}^{2} + C F^{2}]$

$\Rightarrow C A^{2} + B C^{2} = \frac{A B^{2}}{2} + \frac{9}{2} C G^{2} ⟶ 3 [∵ C G : G F = 2 : 1]$

$1 + 2 + 3 \Rightarrow 2 [{A B}^{2} + {B C}^{2} + {C A}^{2}] = \frac{1}{2} [{A B}^{2} + {B C}^{2} + {C A}^{2}] + \frac{9}{2} [{A G}^{2} + {B G}^{2} + {C G}^{2}]$

${A B}^{2} + {B C}^{2} + {C A}^{2} = 3 [{A G}^{2} + {B G}^{2} + {C G}^{2}]$
Hence Proved.

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