Question

If G be the centroid of a triangle ABC, prove that :

$A{B}^2+B{C}^2+C{A}^2=3(G{A}^2+G{B}^2+G{C}^2)$

Answer 1

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, i.e. G = (0, 0).

Centroid of a triangle= $\frac{(x_1+x_2+x_3)}{3}$, $\frac{(y_1+y_2+y_3)}{3}$

Since it is given that the centroid of a triangle G = ( 0,0 )

Hence $\frac{(x_1+x_2+x_3)}{3}$ =0 and $\frac{(y_1+y_2+y_3)}{3}$= 0

$\to$ 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 + y₃² – 2y₂y₃) + (x₁² + x₃² – 2x₁x₃ + y₁² + y₃² – 2y₁y₃)

= (2x₁² + 2x₂² + 2x₃² – 2x₁x₂ – 2x₂x₃ – 2x₁x₃) + (2y₁² + 2y₂² + 2y₃² – 2y₁y₂ – 2y₂y₃ – 2y₁y₃)

= (3x₁² + 3x₂² + 3x₃²) + (3y₁² + 3y₂² + 3y₃²) (From (1))

= 3(x₁² + x₂² + x₃²) + 3(y₁² + y₂² + y₃²) … (2)

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 + y₁² + x₂² + y₂² + x₃² + y₃²)

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

From (2) and (3), we get

AB² + BC² + CA² = 3(GA² + GB² + GC²)