Question

If G is centroid of triangle ABC and O any other point, prove that 3(GA^2+GB^2+GC^2)=BC^2+CA^2+AB^2

Answer 1

There are 2 methods for solving this question:-
First method:-
Let the coordinates of A, B and C be (x1.y1), (x2.y2) and (x3.y3)
           Then coordinates of centroid G will be ((x1­ + x2 + x3)/3, (y1 + y2 + y3)/3 )
           Let it be assumed that the centroid lies on the origin
           ⇒ x1­ + x2 + x3 = 0 and y1 + y2 + y3 = 0
           Let the coordinates of point O be (x, y)
           We have to prove OA2 + OB2 + OC2 = GA2 + GB2 + GC2 + 3GO2
           Now, L.H.S. = OA2 + OB2 + OC2
           (x – x1)2 + (y – y1)2 + (x – x2)2 + (y – y2)2 + (x – x3)2 + (y – y3)2
           = 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32 – 2x (x1 + x2 + x3) – 2y(y1 + y2 + y3)
           = 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32
           R.H.S. = GA2 + GB2 + GC2 + 3GO2
          
           = (x1 – 0)2 + (y1 – 0)2 + (x2 – 0)2 + (y2 – 0)2 + (x3 – 0)2 + (y3 – 0)2 + 3[(x – 0)2 + (y – 0)2]
           = x12 + y12 + x22 + y22 + x32 + y32 + 3(x2 + y2)
           = 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32
           = L.H.S

2.Second method:-

By using cosine rule in 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 and replacing BD = BC/2  we get
   2 AB² + 2 AC² = 4 AD² + BC²         
=>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
Now,on adding the 3 equations that we have got we can prove that
3(GA^2+GB^2+GC^2)=BC^2+CA^2+AB^2

Hope it helps!!