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.
∴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.
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