Let $G_1, G_2, G_3$ be the centroids of the triangular faces $OBC, OCA, OAB$ of a tetrahedron $OABC$. If $V_1$ denote the volume of the tetrahedron $OABC$ and $V_2$ that of the parallelopiped with $OG_1, OG_2, OG_3$ as three concurrent edges, then

$4V_1 =9V_2$

Taking O as the origin, let the position vectors of A, B and C be $\vec a,\vec b$ and $\vec c$ respectively. Then, the position vectors of $G_1, G_2$ and $G_3$ are $\frac{\vec b+\vec c}{3},\frac{\vec c+\vec a}{3}$ and $\frac{\vec a+\vec b}{3}$ respectively.

$∴V_1=\frac{1}{6}[\vec a\,\,\vec b\,\,\vec c]$ and $V_2=[\vec{OG_1}\,\, \vec{OG_2}\,\, \vec{OG_3}]$

$V_2=[\vec{OG_1}\,\, \vec{OG_2}\,\, \vec{OG_3}]$

$⇒V_2=[\frac{\vec b+\vec c}{3}\,\, \frac{\vec c+\vec a}{3}\,\, \frac{\vec a+\vec b}{3}]$

$⇒V_2=\frac{1}{27}\begin{bmatrix}\vec b+\vec c&\vec c+\vec a&\vec a+\vec b\end{bmatrix}$

$⇒V_2=\frac{1}{27}[\vec a\,\,\vec b\,\,\vec c]⇒V_2=\frac{2}{27}×6V_1⇒9V_2=4V_1$