For example, as shown in Figure 4, in quadrilateral $A B C D$, $E$ is the midpoint of $B C$, and $A E$ intersects $B D$ at $F$. If $D F = B F, A F = 2 E F$, then $S_{\triangle A C D}: S_{\triangle A B C}: S_{\triangle A B D}$ $=$ $\qquad$ . ( $S_{\triangle A C D}$ represents the area of $\triangle A C D$, and so on)
Solution: Connect $D E$.
$$
\begin{array}{l}
\text { Let } S_{\triangle X E F}=S . \\
\because A F=2 E F, \\
\therefore S_{\triangle B A F}=2 S_{\triangle B E F}=2 S . \\
\text { Also, } \because D F=B F, \\
\therefore S_{\triangle D E F}=S_{\triangle B E F}=S .
\end{array}
$$

Similarly, $S_{\triangle D E C}=S_{\triangle D B E}=2 S$,
$$
\begin{array}{l}
S_{\triangle A B C}=2 S_{\triangle A B E}=6 S, \\
S_{\triangle A B D}=2 S_{\triangle B A F}=4 S:
\end{array}
$$
$\because E F$ is the midline of $\triangle B D C$,
$$
\begin{array}{l}
\therefore A E \parallel D C . \\
\therefore S_{\triangle A C D}=S_{\triangle E C D}=2 S . \\
\therefore S_{\triangle A C D}: S_{\triangle A B C}: S_{\triangle A B D}=2 S: 6 S: 4 S \\
=1: 3: 2 .
\end{array}
$$