Example 5. The side length of square $ABCD$ is $1$, and there are points $P, Q$ on $AB, AD$ respectively. If the perimeter of $\triangle APQ$ is 2, find $\angle PCQ$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
This is a problem from the 1st (1987) Mathematical Winter Camp for middle school students in our country, and in the 4th (1991) "Zu Chongzhi Cup" Mathematical Competition, it was also used as a contest problem ${ }^{[5]}$. The usual proof method involves the use of congruent shapes. In fact, points $P$ and $Q$ in this example also have a "symmetry," because swapping the positions of $P$ and $Q$ still makes the original problem true. To utilize this symmetry, we can assume
$$
P B=a, Q D=b .
$$

Given that $A B+A D=2$ and the perimeter of $\triangle A P Q$ is 2, it is easy to see that
$$
P Q=a+b .
$$

But $P Q^{2}=A P^{2}+A Q^{2}=(1-a)^{2}+(1-b)^{2}$,
thus from $(1-a)^{2}+(1-b)^{2}=(a+b)^{2}$ we get
$$
a+b=1-a b .
$$

This equation's symmetry with respect to $a$ and $b$ precisely reflects the "symmetric" status of points $P$ and $Q$. Fully exploring its implications can lead to an "elegant" solution.
Let $\angle P C B=\alpha, \angle Q C D=\beta$, it is easy to know
$$
a=\operatorname{tg} \alpha, b=\operatorname{tg} \beta \text {. }
$$

The above equation can be transformed into
$$
1=\frac{a+b}{1-a b}=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\operatorname{tg}(\alpha+\beta) .
$$

Thus, $\alpha+\beta=45^{\circ}$.
Therefore, $\angle P C Q=45^{\circ}$.