Example 1 As shown in Figure 1, given that the area of $\triangle ABC$ is 1, points $D$, $E$, and $F$ are on sides $BC$, $CA$, and $AB$ respectively, with $BD=2DC$, $CE=2EA$, and $AF=2FB$. $AD$, $BE$, and $CF$ intersect pairwise at points $P$, $Q$, and $R$. Find the area of $\triangle PQR$.
(2009, University of Science and Technology of China Independent Admission Examination)
Solve: By applying Ceva's Theorem to $\triangle A D C$ cut by line $B P E$, we get
$$
\begin{array}{l}
\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A}=1 \\
\Rightarrow \frac{A P}{P D} \cdot \frac{2}{3} \cdot \frac{2}{1}=1 \Rightarrow \frac{A P}{P D}=\frac{3}{4} \\
\Rightarrow S_{\triangle N A B}=\frac{3}{7} S_{\triangle N R D}=\frac{3}{7} \times \frac{2}{3} S_{\triangle M B C}=\frac{2}{7} S_{\triangle A B C}
\end{array}
$$

Similarly, $S_{\triangle B Q C}=S_{\triangle M R C}=\frac{2}{7} S_{\triangle M B C}$.
Thus, $S_{\triangle P Q R}=S_{\triangle A R C}-S_{\triangle A P B}-S_{\triangle A Q C}-S_{\triangle A R C}$ $=\frac{1}{7} S_{\triangle A B C}=\frac{1}{7}$.