3. $P$ is a point inside $\triangle A B C$, and line segments $A P D, B P E$, and $C P F$ are drawn such that $D$ is on $B C$, $E$ is on $A C$, and $F$ is on $A B$. Given that $A P=6, B P=9, P D=6, P E=3, C F=20$, find the area of $\triangle A B C$.
(7th AIME Problem)
3. For $\triangle B D P$ and the transversal $C E A$, we have $\frac{B C}{C D} \cdot \frac{D A}{A P} \cdot \frac{P E}{E A}=\frac{B C}{C D} \cdot \frac{12}{6} \cdot \frac{3}{12}=1$, which implies $B D=D C$. For $\triangle C D P$ and the transversal $B F A$, we have $\frac{C B}{B D} \cdot \frac{D A}{A P} \cdot \frac{P F}{F C}=\frac{2}{1} \cdot \frac{2}{1} \cdot \frac{P F}{F C}=1$, which implies $\frac{P F}{F C}=\frac{1}{4}$. Given $C F=20$, we have $C P=15$.

In $\triangle P B C$, using the median length formula $P D^{2}=\frac{1}{2} \sqrt{2\left(P B^{2}+P C^{2}\right)-B C^{2}}$, we get $B C=2 \sqrt{117}$, hence $B D=\sqrt{117}$. Also, $B P^{2}+P D^{2}=6^{2}+9^{2}=(\sqrt{117})^{2}=B D^{2}$, which implies $\angle B P D=90^{\circ}, S_{\triangle P B D}=27, S_{\triangle A B C}=4 S_{\triangle P B D}=108$.