24. In $\triangle A B C$, point $D$ is on side $A C$, and $\frac{A D}{D C}=\frac{1}{2}, E$ is the midpoint of segment $B D$, and line $A E$ intersects $B C$ at point $F$. If $S_{\triangle A B C}=360$, then the area of $\triangle E B F$ is ( ).
(A) 24
(B) 30
(C) 32
(D) 36
(E) 40
24. B.

In $\triangle B C D$, by Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{B F}{F C} \cdot \frac{C A}{A D} \cdot \frac{D E}{E B}=\frac{3 B F}{F C}=1 \\
\Rightarrow \frac{B F}{F C}=\frac{1}{3} \Rightarrow \frac{B F}{B C}=\frac{1}{4} .
\end{array}
$$

Thus, $S_{\triangle E B F}=\frac{B E}{D B} \cdot \frac{B F}{B C} S_{\triangle B C D}$
$$
=\frac{1}{2} \times \frac{1}{4}\left(\frac{2}{3} S_{\triangle A B C}\right)=30 \text {. }
$$