Example 6 As shown in Figure 2-9, in $\triangle A B C$, $D, E$ are points on $B C, C A$ respectively, and $B D: D C=m: 1, C E: E A=n: 1, A D$ intersects $B E$ at $F$, then the area of $\triangle A B F$ is how many times the area of $\triangle A B C$?
(1984 Shanghai Middle School Competition Question)
Extend $CF$ and let it intersect $AB$ at $G$. Apply Ceva's Theorem to $\triangle ABC$ and point $F$, we have
$$
\frac{AG}{GB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = \frac{AG}{GB} \cdot m \cdot n = 1.
$$
$$
\therefore \frac{AG}{GB} = \frac{1}{m \cdot n}, \text{ i.e., } \frac{AB}{BG} = \frac{mn+1}{mn}.
$$

Apply Menelaus' Theorem to $\triangle AGC$ and the transversal $BFE$, we have
$$
\frac{AB}{BG} \cdot \frac{GF}{FC} \cdot \frac{CE}{EA} = \frac{mn+1}{mn} \cdot \frac{GF}{FC} \cdot n = 1, \therefore \quad \frac{GF}{FC} = \frac{m}{mn+1}.
$$

Therefore, $\frac{S_{\triangle ABF}}{S_{\triangle ABC}} = \frac{GF}{GC} = \frac{GF}{FC + GF} = \frac{m}{mn + m + 1}$ is the desired result.