Question

If   G  is the CENTROID of triangle ABC then prove that
AB∧2 +BC∧2+AC∧2=3(GA∧2+GB∧2+GC∧2)  BEST OF LUCK FOR THE EQUATION 

Answer 1

Using cosine rule in a triangle ABC and triangle ABD  (D = mid point of BC, E = mid point of AC, and F = mid point of AB)
cos B = (AB² + BC² - AC² ) / 2 AB * AC
cos B = (AB² + BD² - AD² ) / 2 AB *  AD
   equating both sides, replace BD = BC/2  we get
   2 AB² + 2 AC² = 4 AD² + BC²            write now AD =  GA + GD  = (3/2) GA
                         = 4 (9/4 GA² ) + BC² = 9 GA²+ BC²
  similarly  2 AB² + 2 BC²  =9 GB2  + AC2
           2 BC² + 2 AC² = 9 GC2 + AB2
Add the three equations
   you get answer