Problem - 2454
Let $G$ be the centroid of $\triangle{ABC}$, $L$ be a straight line. Prove that $$GG'=\frac{AA'+BB'+CC'}{3}$$ where $A'$, $B'$, $C'$ and $G'$ are the feet of perpendicular lines from $A$, $B$, $C$, and $G$ to $L$.
As shown, let $M$ be the middle point of $AC$. Then, it is easy to see that $MM'=\frac{AA'+CC'}{2}$. Meanwhile, because $BG=2GM$, we have $GG'=\frac{2MM'+BB'}{3}$. Combing these two, we have $$GG'=\frac{1}{3}\cdot(2MM'+BC')=\frac{1}{3}\cdot(2\cdot\frac{AA'+CC'}{2}+BB')=\frac{AA'+BB'+CC'}{3}$$
