If G be the centroid of a triangle ABC and P be any other point in the plane, prove that $P A^{2} + P B^{2} + P C^{2} = G A^{2} + G B^{2} + G C^{2} + 3 G P^{2}$ .

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Solution

$N o w, L . H . S (x - a)^{2} + (y - b)^{2} + (x - c)^{2} + (y - d)^{2} + (x + a + c)^{2} + (y + b + d)^{2} = (x^{2} - 2 a x + a^{2}) + (y^{2} - 2 b y + b^{2}) + (x^{2} - 2 c x + c^{2}) + (y^{2} - 2 d y + y^{2}) + (x^{2} + a^{2} + c^{2} + 2 a x + 2 a c + 2 c x) + (y^{2} + b^{2} + d^{2} + 2 b y + 2 b d + 2 d y) = 3 (x^{2} + y^{2}) + 2 (a^{2} + b^{2} + c^{2} + d^{2} a b + c d) N o w, R . H . S [(0 - a)^{2} + (0 - b)^{2}] + [(0 - c)^{2} + (0 - d)^{2}] + [(0 + a + c)^{2} + (0 + b + d)^{2}] + 3 [(x - 0)^{2} + (y - 0)^{2}] = a^{2} + b^{2} + c^{2} + d^{2} + (a^{2} + 2 a c + c^{2}) + (b + d^{2} + 2 b d) + 3 (x^{2} + y^{2}) = 3 (x^{2} + y^{2}) + 2 (a^{2} + b^{2} + c^{2} + d^{2} + a c + b a) = L . H . S$
Thus, $P A^{2} + P B^{2} + P C^{2} = G A^{2} + G B^{2} + G C^{2} + 3 G P^{2}$

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