2. Let the volume of tetrahedron $ABCD$ be $V$, $E$ be the midpoint of edge $AD$, and $F$ be on the extension of $AB$ such that $BF = AB$. The plane through points $C, E, F$ intersects $BD$ at $G$. Then the volume of tetrahedron $CDGE$ is $\qquad$
2. $\frac{V}{3}$ In $\triangle A F D$, since point $G$ is the intersection of median $B D$ and $E F$, $\therefore D G=\frac{2}{3} B D$.
$$
\begin{array}{l}
\because D E=\frac{1}{2} A D, \\
\therefore S_{\triangle D G E} \\
=\frac{1}{3} S_{\triangle A B D},
\end{array}
$$

Since $V_{C-D G E}$ and $V_{C-A B D}$ have the same height, the volume of tetrahedron $C-D G E$ is $\frac{V}{3}$.