If $ABC$ and $A'B'C'$ are two triangles and $G$ and $G'$ be their corresponding centroids, prove that $\overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} = 3 \overrightarrow{GG'}$
Let $O$ be the origin.
Let the position vectors of points $A$, $B$ and $C$ be
$\overrightarrow{OA} = \overrightarrow{a}$, $\overrightarrow{OB} = \overrightarrow{b}$ and $\overrightarrow{OC} = \overrightarrow{c}$ respectively.
$G$ is the centroid of $\triangle ABC$.
$\therefore$ $\;$ Position vector of $G = \overrightarrow{OG} = \dfrac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3}$
Let the position vectors of points $A'$, $B'$ and $C'$ be
$\overrightarrow{OA'} = \overrightarrow{a'}$, $\overrightarrow{OB'} = \overrightarrow{b'}$ and $\overrightarrow{OC'} = \overrightarrow{c'}$ respectively.
$G'$ is the centroid of $\triangle A'B'C'$.
$\therefore$ $\;$ Position vector of $G' = \overrightarrow{OG'} = \dfrac{\overrightarrow{a'} + \overrightarrow{b'} + \overrightarrow{c'}}{3}$
Now,
$\overrightarrow{AA'} = \overrightarrow{OA'} - \overrightarrow{OA} = \overrightarrow{a'} - \overrightarrow{a}$
$\overrightarrow{BB'} = \overrightarrow{OB'} - \overrightarrow{OB} = \overrightarrow{b'} - \overrightarrow{b}$
$\overrightarrow{CC'} = \overrightarrow{OC'} - \overrightarrow{OC} = \overrightarrow{c'} - \overrightarrow{c}$
$\begin{aligned}
\therefore \; \overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} & = \overrightarrow{a'} - \overrightarrow{a} + \overrightarrow{b'} - \overrightarrow{b} + \overrightarrow{c'} - \overrightarrow{c} \\\\
& = \left(\overrightarrow{a'} + \overrightarrow{b'} + \overrightarrow{c'}\right) - \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) \\\\
& = 3 \left(\dfrac{\overrightarrow{a'} + \overrightarrow{b'} + \overrightarrow{c'}}{3}\right) - 3 \left(\dfrac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3}\right) \\\\
& = 3 \left(\overrightarrow{OG'} - \overrightarrow{OG}\right) \\\\
& = 3 \overrightarrow{GG'}
\end{aligned}$
Hence proved.