If G be the centroid of a triangle ABC, prove that, `AB^2 + BC^2 + CA^2 = 3(GA^2 + GB^2 + GC^2)`

Explanation

Let `A(x1,y1),B(x2,y2) and C(x3,y3)` be the vertices of `/_\ ABC`


Centroid of `/_\ABC=[(x1+x2+x3​)/3,(y1+y2+y3)/3​]`


So,`x1+x2+x3=0;y1+y2+y3=0`


Squaring on both sides,

`x1^2+x2^2+x3^2+2x1.x2+2x2.x3+2x3.x1=0
...