(20 13? Zhuhai) as shown in the figure, in Rt△ABC, ∠ c = 90, point P is a point on the AC side, and the line segment AP rotates clockwise around point A (point P corresponds.

(1) proves that ∵AP' is obtained by AP rotation,

∴ap=ap′，

∴∠app′=∠ap′p，

∠∠c = 90 ,ap′⊥ab，

∴∠CBP+∠BPC=90，∠ABP+∠AP′p = 90，

∫∠BPC =∠APP' (equal to the vertex angle),

∴∠cbp=∠abp；

(2) Proof: As shown in the figure, the passing point P is PD⊥AB in D,

∠∠CBP =∠ABP，∠C=90，

∴CP=DP，

∵p′e⊥ac，

∴∠eap′+∠ap′e=90，

And ∵∠ pad+∠ EAP' = 90,

∴∠pad=∠ap′e，

In △APD and △ P ′ AE, ∠ pad = ∠ AP ′ e ∠ ADP = ∠ P ′ ea = 90 AP = AP ′,

∴△apd≌△p′ae(aas)，

∴AE=DP，

∴ae=cp；

(3) Solution: CPPE = 32,

∴ Let CP=3k, PE=2k,

AE=CP=3k，AP'=AP=3k+2k=5k，

At rt △ AEP ′′′, p ′ e = (5k) 2? (3k)2=4k，

∠∠c = 90 ,p′e⊥ac，

∴∠CBP+∠BPC=90，∠EP′p+∠EPP′= 90，

∠∠BPC =∠∠EPP' (equal to the vertex angle),

∴∠cbp=∠ep′p，

* CBP =∠ABP，∴∠ABP=∠EP'P，

∠∠BAP′=∠P′EP = 90，

∴△abp′∽△epp′，

∴abp′e=p′ape，

That is AB4k=P'A2k,

p’a = 12ab，

In rt △ ABP ′′′, AB2+P ′ A2 = BP ′ 2,

AB2+ 14AB2=(55)2，

The solution is ab = 10.