Example 5. As shown in the figure, in $\triangle A B C$, $D, E$ are points on $B C, C A$ respectively, $B E, A D$ intersect at $H$, and $\frac{B D}{D C}=\frac{1}{3}, \frac{C E}{E A}=\frac{1}{2}, S_{\triangle A B C}=1$. Find the area of $\triangle A H E$.
In Theorem 4, let
$$
m=\frac{B D}{D C}=\frac{1}{3}, n=\frac{C E}{E A}, p=0
$$
(considering $A, F$ coincide), then
$$
\begin{aligned}
\frac{S_{\triangle A H F}}{S_{\triangle A B C}} & =\frac{1}{\left(1+\frac{1}{3}+\frac{1}{6}\right)\left(1+\frac{1}{2}\right)} \\
& =\frac{4}{9} . \\
\therefore S_{\triangle A H F} & =\frac{4}{9} .
\end{aligned}
$$