4. In $\triangle A B C$, $D$ is on $B C$, $B D: D C=$ $3: 2, E$ is on $A D$, $A E: E D=5: 6$, extend $B E$ to intersect $A C$ at $F$. Then $B E: E F=$ . $\qquad$
$4.9: 2$
Solve using the Angle Bisector Theorem. As shown in the figure, we have
$$
\begin{array}{l}
\frac{A F}{F C}=\frac{A B \cdot \sin \angle 1}{B C \cdot \sin \angle 2} \\
=\frac{A B \cdot \sin \angle 1}{\frac{5}{3} B D \cdot \sin \angle 2}
\end{array}
$$

Thus, $A F=\frac{1}{3} A C$.
$$
\text { Also, } \frac{B E}{E F}=\frac{A B \cdot \sin \alpha}{A F \cdot \sin \beta}
$$

Thus, $B E: E F=9: 2$.