If G be the centroid of a triangle ABC, prove that:
AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2)
We know that centroid of a triangle for (x1 , y1), (x2 , y2) and (x3 , y3) is
G(x, y) = ( 
, 
)
We assume centroid of ∆ABC at origin.
For x=0 and y=0
= 0 and 
= 0
∴ 
= 0 and 
=0
Squaring on both sides, we get
x12 + x22 + x32 + 2x1x2 + 2x2x3 + 2x3x1 = 0 and y12 + y22 + y32 + 2y1y2 + 2y2y3 + 2y3y1 = 0 … (1)
AB2 + BC2 + CA2
= [(x2 – x1)2 + (y2 – y1)2] + [(x3 – x2)2 + (y3 – y2)2] + [(x1 – x3)2 + (y1 – y3)2]
= (x12 + x22 – 2x1x2 + y12 + y22 – 2y1y2)+(x22 + x32 – 2x2x3 + y22 + y32 – 2y2y3)+(x12 + x32 – 2x1x3 + y12 + y32 - 2y1y3)
= (2x12 + 2x22 + 2x32 – 2x1x2 – 2x2x3 – 2x1x3) + (2y12 + 2y22 + 2y32 – 2y1y2 – 2y2y3 – 2y1y3)
= (3x12 + 3x22 + 3x32) + (3y12 + 3y22 + 3y32) …from 1
= 3(x12 + x22 + x32) + 3(y12 + y22 + y32) … (2)
3(GA2 + GB2 + GC2)
= 3 [(x1 – 0)2 + (y1 – 0)2 + (x2 – 0)2 + (y2 – 0)2 + (x3 – 0)2 + (y3 – 0)2]
= 3 (x12 + y12 + x22 + y22 + x32 + y32)
= 3 (x12 + x22 + x32) + 3(y12 + y22 + y32) … (3)
From (2) and (3), we get
AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2)