Centroid Vector Identity

Let $M$ be the midpoint of side $BC$. Since $M$ bisects $BC$, the two vectors $\overrightarrow{MB}$ and $\overrightarrow{MC}$ have equal length and opposite directions, so $\overrightarrow{MB}+\overrightarrow{MC}=\vec{0}$. By the triangle rule from $G$ through $M$, we have $\overrightarrow{GB}+\overrightarrow{GC}=(\overrightarrow{GM}+\overrightarrow{MB})+(\overrightarrow{GM}+\overrightarrow{MC})=2\overrightarrow{GM}$.

By the median property of $\triangle ABC$, the centroid $G$ lies on the median $AM$ and divides it in a $2:1$ ratio from the vertex, meaning $\overrightarrow{GA}=-2\overrightarrow{GM}$.

Adding the two relations yields $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=-2\overrightarrow{GM}+2\overrightarrow{GM}=\vec{0}$. Finally, for any point $O$, writing $\overrightarrow{GA}=\overrightarrow{OA}-\overrightarrow{OG}$ (and similarly for $B$ and $C$) gives $(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})-3\overrightarrow{OG}=\vec{0}$, or $\overrightarrow{OG}=\dfrac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$.