1. As shown in Figure 1, let the area of $\triangle A B C$ be $1, A F=\frac{1}{m} A B, B D=\frac{1}{n} B C, C E=\frac{1}{p} C A$. Then the area of $\triangle D E F$ is . (1984, Xi'an City Junior High School Competition)
From the conditions, we have
$$
\frac{A F}{F B}=\frac{1}{m-1}, \frac{B D}{D C}=\frac{1}{n-1}, \frac{C E}{E A}=\frac{1}{p-1} .
$$

Substituting $\frac{1}{m-1}, \frac{1}{n-1}, \frac{1}{p-1}$ for $m, n, l$ in the theorem, respectively, and simplifying, we get
$$
\begin{aligned}
S_{\triangle D E F}= & \frac{1}{m n p}(m n p-m n-n p-p m+m \\
& +n+p) .
\end{aligned}
$$