If G be the centroid of a triangle ABC, prove that $A B^{2} + B C^{2} + C A^{2} = 3 (G A^{2} + G B^{2} + G C^{2}) .$

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Solution

$L e t A (x 1, y 1), B (x 2, y 2) a n d C (x 3, y 3) b e t h e v e r t i c e s o f △ A B C .$
$W i t h o u t t h e l a w o f g e n e r a l i t y, a s s u m e t h e c e n t r o i d o f t h e$
$△ A B C t o b e a t o r i g i n, i . e ., G = (0, 0) .$
$C e n t r o i d o f △ A B C = [\frac{x 1 + x 2 + x 3}{3}, \frac{y 1 + y 2 + y 3}{3}]$
$∴ x 1 + x 2 + x 3 = 0; y 1 + y 2 + y 3 = 0$
$S q u a r i n g o n b o t h s i d e s,$
${x 1}^{2} + {x 2}^{2} + {x 3}^{2} + 2 x 1. x 2 + 2 x 2. x 3 + 2 x 3. x 1 = 0$
${y 1}^{2} + {y 2}^{2} + {y 3}^{2} + 2 y 1. y 2 + 2 y 2. y 3 + 2 y 3. y 1 = 0 - (i)$
${A B}^{2} + {B C}^{2} + {C A}^{2}$
$= [{(x 2 - x 1)}^{2} + {(y 2 - y 1)}^{2}] + [{(x 3 - x 2)}^{2} + {(y 3 - y 2)}^{2}] + [{(x 1 - x 3)}^{2} + {(y 1 - y 3)}^{2}]$
$= ({x 1}^{2} + {x 2}^{2} - 2 x 1 x 2 + {y 1}^{2} + {y 2}^{2} - 2 y 1 y 2) + ({x 1}^{2} + {x 3}^{2} - 2 x 1 x 3 + {y 1}^{2} + {y 3}^{2} - 2 y 2 y 3)$
$+ ({x 1}^{2} + {x 3}^{2} - 2 x 1 x 3 + {y 1}^{2} + {y 3}^{2} - 2 y 1 y 3)$
$= (2 {x 1}^{2} + 2 {x 2}^{2} + 2 {x 3}^{2} - 2 x 1 x 2 - 2 x 2 x 3 - 2 x 1 x 3) +$
$(2 {y 1}^{2} + 2 {y 2}^{2} + 2 {y 3}^{2} - y 1 y 2 - 2 y 2 y 3 - 2 y 1 y 3)$
$= (3 {x 1}^{2} + 3 {x 2}^{2} + 3 {x 3}^{2}) + (3 {y 1}^{2} + 3 {y 2}^{2} + 3 {y 3}^{2})$
$= 3 ({x 1}^{2} + {x 2}^{2} + {x 3}^{2}) + 3 ({y 1}^{2} + {y 2}^{2} + {y 3}^{2}) - (i i)$
$3 (G A^{2} + G B^{2} + G C^{2})$
$= 3 [{(x 1 - 0)}^{2} + {(y 1 - 0)}^{2} + {(x 2 - 0)}^{2} + {(y 2 - 0)}^{2} + {(x 3 - 0)}^{2} + {(y 3 - 0)}^{2}]$
$= 3 [{x 1}^{2} + {y 1}^{2} + {x 2}^{2} + {y 2}^{2} + x 3^{2} + {y 3}^{2}]$
$= 3 ({x 1}^{2} + {x 2}^{2} + {x 3}^{2}) + 3 ({y 1}^{2} + {y 2}^{2} + {y 3}^{2}) - (i i i)$
$f r o m (i i) & (i i i)$
${A B}^{2} + {B C}^{2} + {C A}^{2} = 3 ({G A}^{2} + {G B}^{2} + {G C}^{2})$

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