Example 20 In the right angle $\angle P O Q$, there is a moving point $C$. Try to find a point $A$ on $O P$ and a point $B$ on $O Q$ such that $B C + C A$ equals a fixed length $l$, and the area of quadrilateral $A O B C$ is maximized.
Observe the structure of the given figure; it gives an incomplete feeling. By completing the symmetrical figure as shown in Figure 1.4.22, we get the octagon $A C^{\prime} B^{\prime} C^{\prime} A^{\prime} C^{\prime \prime} B C$. Thus, when the area of the octagon is maximized, the area of the quadrilateral $A O B C$ is also maximized. Since $B C + C A$ is a fixed length $l$, this implies that the perimeter of the octagon is a constant $4 l$. We know that among octagons with a fixed perimeter, the regular octagon has the maximum area. Therefore, when $C$ is on the angle bisector of $\angle P O Q$, and $O A = O B = O C = \frac{1}{4 \sin 22.5^{\circ}}$, the area of the quadrilateral $A O B C$ is maximized.