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


2. Take the lines $A B, C D$ as coordinate axes, establish a coordinate system, and solve using the analytical method.

Let $P(a, 0), Q(0, b)$, then $k_{C P}=\frac{1}{1-a}, k_{C Q}=1-b$.
From the perimeter of $\triangle A P Q$ being 2, we get $a b=2(a+b)-2$.
Since $\operatorname{tg} \angle P C Q=\frac{2-(a+b)}{2-(a+b)}=1$, and $\angle P C Q$ is an acute angle, thus $\angle P C Q=45^{\circ}$.