Example 2. Given that $D, F$ are points on the sides $A B$, $A C$ of $\triangle A B C$ respectively, and $A D: D B=C F: F A=2: 3$. Connect $D F$ to intersect the extension of side $B C$ at point $E$. Then, $E F: F D=$ $\qquad$
(3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
Given Figure 3, line $D F E$ intersects $\triangle A B C$, we have
$$
\begin{aligned}
& \frac{C F}{F A} \cdot \frac{A D}{D B} \cdot \frac{B E}{E C} \\
= & 1 . \\
& \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{B E}{E C}=1, \\
& \frac{B E}{E C}=\frac{9}{4}, \frac{B C}{C E}=\frac{5}{4} .
\end{aligned}
$$

Line $A F C$ intersects $\triangle D B E$,
$$
\begin{aligned}
\frac{E F}{F D} \cdot \frac{A D}{A B} \cdot \frac{B C}{C E}=1 . \\
\frac{E F}{F D} \cdot \frac{2}{5} \cdot \frac{5}{4}=1 .
\end{aligned}
$$

Therefore, $E F: F D=2: 1$.