If G be the centroid of a triangle ABC, prove that AB2+ BC2 + AC2 = 3(GA2 + GB2 + GC2).

Solution:

Let A (x1, y1), B(x2, y2), and C(x3, y3) be the vertices of ∆ABC.

Without the loss of Generality, assume the centroid of the ΔABC to be at the origin, G = (0, 0).

 Centroid of the ΔABC = [(x1+x2+x3)/3, (y1+y2+y3)/3]

=> x1+x2+x3 = 0 and y1+y2+y3 = 0

Squaring above 2 equations, we get

x12+x22 + x32+ 2x1x2 + 2x2x3 + 2x1x3 = 0 and 

y12+y22 + y32+ 2y1y2 + 2y2y3 + 2y1y3 = 0 …(i)

AB2+ BC2 + AC2 = (x2-x1)2 + (y2-y1)2 + (x3-x2)2 + (y3-y2)2 + (x3-x1)2 + (y3-y1)2

= x12 + x22 – 2x1x2 + y12 + y22 – 2y1y2 + x32 + x22 – 2x3x2 + y32 + y22 – 2y3y2 + x12 + x32 – 2x1x3 + y12 + y32 – 2y1y3

= 2x12 + 2x22 + 2x32– 2x1x2 – 2x3x2 – 2x1x3+ 2y12 + 2y22 + 2y32 – 2y1y2 – 2y3y2 – 2y1y3

= 3x12 + 3x22 + 3x32 + 3y12 + 3y22 + 3y32

= 3(x12 + x22 + x32) + 3(y12 + y22 + y32)…(ii)

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 + x22 + x32) + 3(y12 + y22 + y32)…(iii)

From (ii) and (iii), we get 

AB2+ BC2 + AC2 = 3(GA2 + GB2 + GC2)

Hence proved.

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