7. $\frac{V}{3}$ Detailed explanation: As shown in the figure, $B$ is the midpoint of $A F$, and $E$ is the midpoint of edge $A D$. According to Menelaus' theorem, $D G=\frac{2}{3} D B$. Therefore, $\frac{V_{E-C D G}}{V_{A-B C D}}=\frac{D G \cdot D C \cdot D E}{D B \cdot D C \cdot D A}=\frac{1}{3}$, so $V_{E-C D G}=\frac{1}{3} V_{A-B C D}=\frac{1}{3} V$.
7. Let the volume of tetrahedron $ABCD$ be $V$, $E$ be the midpoint of edge $AD$, and point $F$ be on the extension of $AB$ such that $BF = AB$. The plane through points $C$, $E$, and $F$ intersects $BD$ at $G$. Then the volume of tetrahedron $CDGE$ is equal to $\qquad$.