5. Square $A B C D$ has a side length of $1, A B, A D$ each have a point $P, Q$. If the perimeter of $\triangle A P Q$ is 2, find the degree measure of $\angle P C Q$.

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Let $\angle B C P=\alpha, \angle D C Q=\beta$.
$$
\begin{aligned}
\because B P & =\tan \alpha, Q D=\tan \beta, \\
A P & =1-\tan \alpha, A Q=1-\tan \beta . \\
\therefore P Q & =2-A P-A Q=\tan \alpha+\tan \beta .
\end{aligned}
$$

From $A P^{2}+A Q^{2}=P Q^{2}$, we know
$$
(1-\tan \alpha)^{2}+(1-\tan \beta)^{2}=(\tan \alpha+\tan \beta)^{2} \text {, }
$$

Simplifying, we get $\tan(\alpha+\beta)=1$.
$$
\text { Also, } \because 0 \leqslant \alpha+\beta<90^{\circ}, \therefore \alpha+\beta=45^{\circ} \text {. }
$$

Thus, $\angle P C Q=90^{\circ}-\alpha-\beta=45^{\circ}$.