Given the height 455 as shown in Figure 3, the lines $AB$ and $AC$ are tangent to the circle $\odot O$ at points $B$ and $C$ respectively. $P$ is a moving point on the arc $\overparen{BC}$, $PD \perp AB$ at point $D$, $PE \perp AC$ at point $E$, and the lines $DE$ and $BC$ are connected. If the radius of $\odot O$ is 5, and the chord $BC$ is 8 units long, find the maximum area of $\triangle PDE$.
Solve As shown in Figure 3, connect $O B, O C$. Then $O B \perp A B, O C \perp A C$.
Draw radius $O N \perp B C$, with the foot of the perpendicular at $M$.
Then $N$ is the midpoint of arc $\overparen{B C}$.
Thus, $B M=C M=\frac{1}{2} B C=4$.
By the Pythagorean theorem,
$O M=\sqrt{O B^{2}-B M^{2}}=3$.
Furthermore, $M N=O N-O M=2$.
In $\triangle O B C$, using the area formula we get
$\frac{1}{2} O B \cdot O C \sin \angle B O C=S_{\triangle O B C}=\frac{1}{2} B C \cdot O M$
$\Rightarrow \sin \angle B O C=\frac{B C \cdot O M}{O B \cdot O C}=\frac{24}{25}$.
Since $\angle D P E=180^{\circ}-\angle A=\angle B O C$, then $\sin \angle D P E=\frac{24}{25}$.
Draw $P F \perp B C$ at point $F$, and connect $P B, P C$.
By the inscribed angle theorem, we know $\angle D B P=\angle P C F$.
Combining $\angle P D B=90^{\circ}=\angle P F C$, we get
$\triangle P D B \backsim \triangle P F C \Rightarrow \frac{P D}{P F}=\frac{P B}{P C}$.
Similarly, $\frac{P E}{P F}=\frac{P C}{P B}$.
Thus, $P D \cdot P E=P F^{2}$.
Then $S_{\triangle O B C}=\frac{1}{2} P D \cdot P E \sin \angle D P E=\frac{12}{25} P F^{2}$.
Note that, among all the distances from points on the arc of a segment to the chord, the height of the segment is the longest.
Therefore, $P F_{\max }=M N=2$.
Hence, the maximum area of $\triangle P D E$ is $\frac{48}{25}$.