Given a square $ABCD$, $P$ is a moving point on the plane, satisfying $\angle APB = \angle CPD$. Find the locus of point $P$.

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The above text translated into English, retaining the original text's line breaks and format.
Let the center of the square be at the origin,
$$
\begin{array}{l}
A(-1,1), B(-1,-1), C(1,-1), \\
D(1,1), P(x, y) .
\end{array}
$$

By the angle formula and $\angle A P B=\angle C P D$, we get
$$
\begin{array}{l}
\left|\frac{\frac{y+1}{x+1}-\frac{y-1}{x+1}}{1+\frac{y+1}{x+1} \cdot \frac{y-1}{x+1}}\right|=\left|\frac{\frac{y-1}{x-1}-\frac{y+1}{x-1}}{1+\frac{y-1}{x-1} \cdot \frac{y+1}{x-1}}\right| \\
\Rightarrow\left|\frac{2(x+1)}{(x+1)^{2}+y^{2}-1}\right|=\left|\frac{-2(x-1)}{(x-1)^{2}+y^{2}-1}\right| \\
\Rightarrow \frac{x+1}{(x+1)^{2}+y^{2}-1}=\frac{x-1}{(x-1)^{2}+y^{2}-1} \text { or } \\
\frac{x+1}{(x+1)^{2}+y^{2}-1}=\frac{1-x}{(x-1)^{2}+y^{2}-1} .
\end{array}
$$

If $x=0$, it clearly satisfies the condition.
If $x \neq 0$, by the ratio theorem, we get
$$
\begin{array}{l}
\frac{x+1}{(x+1)^{2}+y^{2}-1}=\frac{x-1}{(x-1)^{2}+y^{2}-1} \\
=\frac{(x+1)-(x-1)}{\left[(x+1)^{2}+y^{2}-1\right]-\left[(x-1)^{2}+y^{2}-1\right]}=\frac{1}{2 x}
\end{array}
$$

or
$$
\begin{array}{l}
\frac{x+1}{(x+1)^{2}+y^{2}-1}=\frac{1-x}{(x-1)^{2}+y^{2}-1} \\
=\frac{(x+1)-(1-x)}{\left[(x+1)^{2}+y^{2}-1\right]-\left[(x-1)^{2}+y^{2}-1\right]}=\frac{1}{2},
\end{array}
$$

which means $y= \pm x$ or $x^{2}+y^{2}=2$.
Upon verification, the locus that satisfies the condition $\angle A P B=\angle C P D$ is two arcs of a circle plus a straight line and four rays (as shown in Figure 1), with the equations
$$
\begin{array}{l}
x=0, y= \pm x(x>1 \text { or } x<-1), \\
x^{2}+y^{2}=2(-1<x<1) .
\end{array}
$$
(Jin Lei, Xian Jiaotong University Affiliated High School Qujiang Campus, 710043)