Prove the centroid coordinate formula

Question

How to proof that the coordinate of the centroid of a triangle ABC is given by $\frac{A+B+C}{3}$ using vectors?

Answer 1

Let $G$ be the centroid of $\triangle ABC$ and let $M$ be the midpoint of the side $BC$. Now since one has $AG:GM=2:1$ (do you need the proof?), one has $$\vec{OG}=\frac{1\times \vec{OA}+2\times \vec{OM}}{2+1}=\frac{\vec{OA}+2\times \frac{\vec{OB}+\vec{OC}}{2}}{3}=\frac{\vec{OA}+\vec{OB}+\vec{OC}}{3}.$$

Answer 2

Let G be the centroid of ABC. Then $\vec{GA}+\vec{GB}+\vec{GC}= \vec{0}$. In a frame, $x_{\vec{GA}}+x_\vec{GB}+x_{\vec{GC}}= 0$. Hence $x_{A}+x_{B}+x_{C} = 3 x_{G}$, and you can do the same for other coordinates!