We know that centroid of a triangle for (x1 , y1), (x2 , y2) and (x3 , y3) is
G(x,y) = \((\frac{x_1+x_2+x_3}3,\frac{y_1+y_2+y_3}3)\)
We assume centroid of ∆ABC at origin.
For x = 0 and y = 0
\(\frac{x_1+x_2+x_3}3=0\, and\,\frac{y_1+y_2+y_3}3\,=\,0\)
∴ x1 + x2 + x3 = 0 and y1 + y2 + y3 = 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 + y3 2 – 2y2y3)+(x12 + x32 – 2x1x3 + y12 + y32 - 2y1y3)
From (2) and (3), we get
AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2
