4. As shown in Figure 1, points $D$ and $E$ are on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at point $O$. The areas of $\triangle OBE$, $\triangle OBC$, and $\triangle OCD$ are $15$, $30$, and $24$, respectively. Then $AE: BE$ $=(\quad)$.
(A) $5: 2$
(B) $2: 1$
(C) $5: 4$
(D) $9: 5$
4. B.

Notice,
$$
\begin{array}{l}
\frac{B O}{O D}=\frac{S_{\triangle O B C}}{S_{\triangle O C D}}=\frac{5}{4}, \frac{C O}{O E}=\frac{S_{\triangle O B C}}{S_{\triangle O B E}}=2, \\
\frac{E B}{B A} \cdot \frac{A D}{D C} \cdot \frac{C O}{O E}=1, \\
\frac{D C}{C A} \cdot \frac{A E}{E B} \cdot \frac{B O}{O D}=1 . \\
\text { Let } \frac{A E}{E B}=x, \frac{A D}{D C}=y .
\end{array}
$$

Equations (1) and (2) can be transformed into
$$
\left\{\begin{array} { l } 
{ \frac { 1 } { 1 + x } \cdot y \cdot 2 = 1 , } \\
{ \frac { 1 } { 1 + y } \cdot x \cdot \frac { 5 } { 4 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=2, \\
y=\frac{3}{2}
\end{array} \Rightarrow \frac{A E}{E B}=2 .\right.\right.
$$