(1)证明:∵Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的, ∴AC=AC′,AB=AB′,∠CAB=∠C′AB′, ∴∠CAC′=∠BAB′, ∴∠ABB′=∠AB′B=∠ACC′=∠AC′C, ∴∠ACC′=∠ABB′, 又∵∠AEC=∠FEB, ∴△ACE∽△FBE. (2)解:当β=2α时,△ACE≌△FBE.在△ACC′中, ∵AC=AC′, ∴∠ACC′===90°﹣α, 在Rt△ABC中,∠ACC′+∠BCE=90°,即90°﹣α+∠BCE=90°, ∴∠BCE=α, ∵∠ABC=α, ∴∠ABC=∠BCE, ∴CE=BE,由(1)知:△ACE∽△FBE, ∴∠BEF=∠CEA,∠FBE=∠ACE, 又∵CE=BE, ∴△ACE≌△FBE. |