23.(11分)如图7-2-15,在$△ ABC$中,已知$∠ ABC = 90°$,$AB = BC$,$BO ⊥ AC$,垂足为点$O$,点$P$,$D$分别在直线$AC$和直线$BC$上,且$PB = PD$,$DE ⊥ AC$,垂足为点$E$.
(1)求证:$△ BPO ≌ △ PDE$.
下面的分析思路是:由结论开始,逐步按层次向前探寻“结论”成立的“条件”.请将分析过程补充完整,并写出推理过程.


(2)若$BP$平分$∠ ABO$,求证:$AP = CD$.
(3)若点$P$运动到$OC$的中点$P'$时,如图7-2-16所示,满足题中条件的点$D$也随之在直线$BC$上运动到点$D'$,请直接写出$CD'$与$AP'$的数量关系.

23. (1)$∠ BOP = ∠ PED$；$PB = PD$；$∠ C = ∠ 1 = 45°, ∠ PBD = ∠ 2.$ 证明过程略.
(2) 证明:由(1)可以得到$∠ 3 = ∠ 4$，$BP$平分$∠ ABO$，
$\therefore ∠ ABP = ∠ 3 = ∠ 4.$
又$\because ∠ A = ∠ C = 45°$，$PB = PD$，
$\therefore △ ABP ≌ △ CPD (\mathrm{AAS}).$
$\therefore AP = CD.$
(3) $CD'$与$AP'$的数量关系是$CD' = \frac{\sqrt{2}}{3}AP'$.
提示:设$OP' = P'C = x$，则
$AO = BO = CO = 2x$，$AP' = 3x.$
由(1)知$△ BP'O ≌ △ P'D'E.$
$\therefore D'E = P'O = x$，$P'E = BO = 2x.$
$\therefore CE = x.$
又$\because ∠ E = 90°,$
$\therefore$ 由勾股定理可得$CD' = \sqrt{2}x.$
$\therefore CD'$与$AP'$的数量关系是$CD' = \frac{\sqrt{2}}{3}AP'.$