Example 7 As shown in Figure 7, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3$, the distance from point $P$ to line $A N$ is $P D=12, A D=$
30, and a line is drawn through $P$ intersecting
$A N$ and $A M$ at points
$B$ and $C$ respectively. Find the minimum
value of the area of $\triangle A B C$. ${ }^{[7]}$
Solve As shown in Figure 7, it can be proven: when the line moves to $B_{0} P=P C_{0}$, the area of $\triangle A B C$ is minimized.
Draw $C_{0} Q / / A N$, intersecting $C B$ at point $Q$.
Thus, $\triangle P C_{0} Q \cong \triangle P B_{0} B$.
Also, $S_{\triangle P C_{0} C} \geqslant S_{\triangle P C_{0} Q}=S_{\triangle P B_{0} B}$, then
$S_{\triangle \triangle B C} \geqslant S_{\triangle A B_{0} C_{0}}$.
Next, we find the area of $\triangle A B_{0} C_{0}$.
As shown in Figure 7, draw the altitude $C_{0} E$ from point $C_{0}$ to side $A B$, with $E$ as the foot of the perpendicular.
Since $C_{0} E / / P D$, and $C_{0} P=P B_{0}$, we get
$C_{0} E=2 P D=24$.
In the right triangle $\triangle C_{0} A E$,
$\tan \angle M A N=\frac{C_{0} E}{A E}=3$.
Therefore, $A E=8$.
Then $E D=A D-A E=B_{0} D$.
Hence $\left(S_{\triangle I B C}\right)_{\text {min }}=S_{\triangle M B_{0} C_{0}}=\frac{1}{2} A B_{0} \cdot C_{0} E$
$=\frac{1}{2} \times 52 \times 24=624$.