Example 4 Refer to Figure 1.13.3, in $\triangle A B C$, $A D, B E, C F$ intersect at $P, A P=P D=6$, $B P=9, P E=3, C F=20$, find $S_{\triangle A B C}$.
Let $P$ be the centroid of points $A, B, C$, and $m_{A}=a$. Then, by properties 2 and 3, $P$ is the centroid of points $A$ and $D$, so $m_{D}=a, m_{P}=2a$. Similarly, $P$ is also the centroid of points $B$ and $E$, so $m_{B}+m_{E}=m_{P}=2a$, and $B P \cdot m_{B}=P E \cdot m_{E}$. Solving this, we get $m_{B}=\frac{a}{2}, m_{E}=\frac{3a}{2}$. By property 3, $F$ is the centroid of points $A$ and $B$, so
$$
m_{F}=m_{A}+m_{B}=\frac{3a}{2}, B F: F A=m_{A}: m_{B}=2: 1
$$

Thus,
$$
P F: P C=m_{C}: m_{F}=\left(m_{E}-m_{A}\right): m_{F}=1: 3, P F=20 \cdot \frac{1}{4}=5
$$

If we set $F A=x, F B=2x$, in $\triangle A B P$, by Stewart's Theorem (refer to Example 15 in the first chapter of this section), we have
$$
5^{2}=P F^{2}=\frac{6^{2} \cdot 2x+9^{2} \cdot x}{2x+x}-2x^{2}
$$

Solving this, we get $x=\sqrt{13}$. In $\triangle A B P$, it is easy to see that $\angle A P B=90^{\circ}$, so $S_{\triangle A P B}=\frac{1}{2} \cdot 9 \cdot 6=27$, thus
$$
S_{\triangle A B C}=\frac{C F}{F P} \cdot S_{\triangle A P B}=\frac{20}{5} \cdot 27=108
$$