If G be the centroid of …

Here,G is the centroid of a triangle ABC.
Let A(b, c), B(0, 0) and C(a, 0) be the coordinates of {tex}\Delta{/tex}ABC then coordinates of centroid are {tex}G\left[ {\frac{{a + b+0}}{3},\frac{c+0+0}{3}} \right]{/tex}
To prove:-
(AB)² + (BC)² + (CA)² = 3(GA² + GB² + GC²)
Consider : L.H.S.
=(AB)² +( BC)² + (CA)²
= b² + c² + a² + (a - b)² + c²
= b² + c² + a² + a² + b² - 2ab + c²
= 2a² + 2b² + 2c² - 2ab
Consider : R.H.S.
=3(GA² + GB² + GC²)
{tex}=3\left[ {{{\left( {\frac{{a + b}}{3} - b} \right)}^2} + {{\left( {c - \frac{c}{3}} \right)}^2} + {{\left( {\frac{{a + b}}{3}} \right)}^2}} \right.{/tex}{tex}\left. { + {{\left( {\frac{c}{3}} \right)}^2} + \left( {\frac{{a + b}}{3} - a} \right) + {{\left( {\frac{c}{3}} \right)}^2}} \right]{/tex}

{tex} = 3{\left[ {{{\left( {\frac{{a - 2b}}{3}} \right)}^2} + {{\left( {\frac{{2c}}{3}} \right)}^2} + \left( {\frac{{a + b}}{3}} \right)} \right.^2}{/tex} {tex}\left. { + {{\left( {\frac{c}{3}} \right)}^2} + {{\left( {\frac{{b - 2a}}{3}} \right)}^2} + {{\left( {\frac{c}{3}} \right)}^2}} \right]{/tex}
{tex} = 3\left[ {\frac{{{a^2} + 4{b^2} - 4ab}}{9} + \frac{{4{c^2}}}{9} + \frac{{{a^2} + {b^2} + 2ab}}{9}} \right.{/tex} {tex}\left. { + {{\frac{c^2}{9}}} + \frac{{{b^2} + 4{a^2} - 4ab}}{9} + {{\frac{c^2}{9}}}} \right]{/tex}
{tex}= 3 \left[ {\frac{{{a^2} + 4{b^2} - 4ab + 4{c^2} + {a^2} + {b^2} + 2ab + {c^2} + {b^2} + 4{a^2} - 4ab + {c^2}}}{9}} \right]{/tex}
{tex} = 3\left[ {\frac{{6{a^2} + 6{b^2} + 6{c^2} - 6ab}}{9}} \right]{/tex}
{tex} = 3 \times 3\left[ {\frac{{2{a^2} + 2{b^2} + 2{c^2} - 2ab}}{9}} \right]{/tex}
= 2a² + 2b² + 2c² - 2ab
L.H.S. = R.H.S.
Therefore, (AB)² + (BC)² + (CA)² = 3(GA² + GB² + GC²)