TriangleCenter Basic

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 $L$.

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