A student asks,

- prove for triangle ABC with centroid G, and any point M,

- |MA|^2 + |MB|^2 + |MC|^2 = 3|MG|^2 + (1/3)(|AB|^2 + |BC|^2 + |CA|^2)

MathHelp replies,

- Let M=(m
_{1},m_{2},m_{3}), - let A=(a
_{1},a_{2},a_{3}), - let B=(b
_{1},b_{2},b_{3}), - let C=(c
_{1},c_{2},c_{3}), and - let G=(g
_{1},g_{2},g_{3}).

Notice in the statement to be proved that all lengths (distances) are squared, and g_{i} depends only on a_{i}, b_{i}, and c_{i}, so the statement is true of each dimension independently. Then it's a matter of proving the identity

- (m-a)
^{2}+ (m-b)^{2}+ (m-c)^{2}= 3(m-g)^{2}+ ((a-b)^{2}+ (b-c)^{2}+ (c-a)^{2})/3

To do this, start by expressing g in terms of the a, b, and c, expand both sides, and collect terms.

- 3(m-g)
^{2}+ ((a-b)^{2}+ (b-c)^{2}+ (c-a)^{2})/3 = - ((3m-a-b-c)
^{2}+ (a-b)^{2}+ (b-c)^{2}+ (c-a)^{2})/3 = - (9m
^{2}+a^{2}+b^{2}+c^{2}-6ma-6mb-6mc+2bc+2ca+2ab + a^{2}-2ab+b^{2}+ b^{2}-2bc+c^{2}+ c^{2}-2ca+a^{2})/3 = - (9m
^{2}+3a^{2}+3b^{2}+3c^{2}-6ma-6mb-6mc)/3 = - (m-a)
^{2}+ (m-b)^{2}+ (m-c)^{2}

Brute force, yes, but it works.