24. D.

As shown in Figure 8.
In $\triangle A B C$, since $A G$ bisects $\angle B A C$, we have $\frac{C G}{B G}=\frac{A C}{A B}=\frac{1}{5}$.
Also, $S_{\triangle A B C}=120$, then $S_{\triangle A B G}=100$.
Since $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively, we know $D E \Perp \frac{1}{2} B C$.
Notice that, $\triangle A F D \backsim \triangle A G B$.
$$
\begin{array}{l}
\text { Then } \frac{S_{\triangle A F D}}{S_{\triangle A G B}}=\frac{1}{4} \Rightarrow S_{\triangle A F D}=25 \\
\Rightarrow S_{\text {quadrilateral } F D B G}=100-25=75 .
\end{array}
$$
24. In $\triangle A B C$, $A B=50, A C=10, S_{\triangle A B C}$ $=120, D, E$ are the midpoints of sides $A B, A C$ respectively, and the angle bisector of $\angle B A C$ intersects $D E, B C$ at points $F, G$. Then the area of quadrilateral $F D B G$ is ( ).
(A) 60
(B) 65
(C) 70
(D) 75
(E) 80