Traditional Culture Encyclopedia - Hotel accommodation - (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.

(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.