(1) 若 $\mathrm{CG}$ 为 $\angle \mathrm{DCF}$ 的平分线. 请判断 $\mathrm{BP}$ 与 $\mathrm{CP}$ 的数量关系, 并 证明;
(2) 若 $\mathrm{AB}=3, \triangle \mathrm{ABP} \cong \triangle \mathrm{CEP}$, 求 $\mathrm{BP}$ 的长.
【答案】 解: (1) $\mathrm{BP}=\mathrm{CP}$, 理由如下:
$\because \mathrm{CG}$ 为 $\angle \mathrm{DCF}$ 的平分线,
\begin{aligned} & \therefore \angle \mathrm{DCG}=\angle \mathrm{FCG}=45^{\circ}, \\ & \therefore \angle \mathrm{PCE}=45^{\circ}, \\ & \because \mathrm{CG} \perp \mathrm{AP}, \\ & \therefore \angle \mathrm{E}=\angle \mathrm{B}=90^{\circ}, \\ & \therefore \angle \mathrm{CPE}=45^{\circ}=\angle \mathrm{APB}, \\ & \therefore \angle \mathrm{BAP}=\angle \mathrm{APB}=45^{\circ}, \\ & \therefore \mathrm{AB}=\mathrm{BP}, \end{aligned}
\begin{aligned} & \because \mathrm{AB}=\frac{1}{2} \mathrm{BC}, \\ & \therefore \mathrm{BC}=2 \mathrm{AB}, \\ & \therefore \mathrm{BP}=\mathrm{PC} ; \\ & \end{aligned}

(2)
\begin{aligned} &\because \triangle \mathrm{ABP} \cong \triangle \mathrm{CEP}, \\ & \therefore \mathrm{AP}=\mathrm{CP}, \\ & \because \mathrm{AB}=3, \\ & \because \mathrm{BC}=2 \mathrm{AB}=6, \\ & \because \mathrm{AP}=\mathrm{AB}{ }^2+\mathrm{BP}^2, \\ & \therefore(6-\mathrm{BP})^2=9+\mathrm{BP}^2, \\ & \therefore \mathrm{BP}=\frac{9}{4} . \end{aligned}