(1) 操作判断

(2) 过移探究
①如㚵 4 , 当点 $P$ 与点 $C$ 重合时, 连接 $D B$, 判断四迄形 $A B D C$ 的形状, 并说明理由:
② 当点 $P$ 与点 $A$, 点 $C$ 都不重合时, 试猜想 $D C$ 与 $B C$ 的位置关系, 并利用图 2 让明你的猜想;
(3) 拓展应用

\begin{aligned} & \therefore A D=A B, \angle B A D=\angle A B C=90^{\circ}, \\ & \therefore A D / / B C, A D=A B=B C, \end{aligned}
$\therefore$ 四边形 $A B C D$ 是平行四边形,
$$\because \angle A B C=90^{\circ}, A B=A C \text {, }$$
$\therefore$ 四边形 $A B C D$ 是正方形;

(2) ①四边形 $A B D C$ 是平行四边形.

\begin{aligned} & \therefore C B=C D, \angle B C D=90^{\circ}, \\ & \because \angle C B A=90^{\circ}, B A=B C, \\ & \therefore \angle B C D=\angle A B C, A B=C D, \\ & \therefore A B / / C D, \end{aligned}
$\therefore$ 四边形 $A B D C$ 是平行四边形;

②猜想: $D C \perp B C$.

\begin{aligned} & \because \angle A B C=90^{\circ}, B A=B C, \\ & \therefore \angle B A C=\angle B C A=45^{\circ}, \\ & \therefore \angle B A C=\angle A E P=45^{\circ}, \\ & \therefore A P=E P, \end{aligned}

$\because$ 将线段 $B P$ 绕点 $P$ 逆时针旋转 $90^{\circ}$ 得到 $P D$,
\begin{aligned} & \therefore P D=P B, \\ & \because \angle A P E=\angle B P I=90^{\circ}, \\ & \therefore \angle A P E+\angle E P B=\angle B P D+\angle E P B, \\ & \therefore \angle A P B=\angle E P D, \end{aligned}

\begin{aligned} & \left\{\begin{array}{l} \mathrm{PA}=\mathrm{PE} \\ \angle \mathrm{APB}=\angle \mathrm{EPD}, \\ \mathrm{PB}=\mathrm{PD} \end{array}\right. \\ & \therefore \triangle A P B \cong \triangle E P D(S A S), \\ & \therefore \angle \mathrm{PAE}=\angle \mathrm{PED}=45^{\circ}, A B=E D, \\ & \therefore \angle A E D=\angle A E P+\angle P E D=90^{\circ}, \\ & \therefore \angle A E D=\angle A B C=90^{\circ}, \\ & \therefore E I) / / B C, \\ & \because A B=B C, \\ & \therefore E D=B C, \end{aligned}

$\therefore$ 四边形 $E B C D$ 是平行四边形,
$$\because \angle A B C=90^{\circ} \text {, }$$
$\therefore$ 四边形 $E B C D$ 是矩形,
\begin{aligned} & \therefore \angle B C D=90^{\circ} \text {, } \\ & \therefore D C \perp B C \text {; } \\ & \text { (3) } C D=3 \sqrt{2} \pm 4 \text {. } \\ & \because \angle B A C=45^{\circ}, P E \perp A C \text {, } \\ & \therefore \triangle E P A \text { 是等腰直角三角形, } \\ & \therefore A E=\sqrt{2} A P \\ & \end{aligned}

H (2) 中② 可知, 四边形 $E B C D$ 是的形, $A B=A E-B E=A E-C D=\sqrt{2} A P-C D$,
\begin{aligned} & \therefore 4=3 \sqrt{2}-C D, \\ & \therefore C D=3 \sqrt{2}-4 . \end{aligned}

\begin{aligned} & \therefore 4=(D)-3 \sqrt{2}, \\ & \therefore C D=3 \sqrt{2}+4 . \end{aligned}