2)∠AEB不变为45°.理由如下:法一:过点A作AH⊥BE垂足为H,作AG⊥CE交CE延长线于G,先证∠ACF=∠ABD,得△BAH≌△CAG(AAS)∴AH=AG,而AH⊥EB,AG⊥EG,∴EA平分∠BEF,∴∠BEA=0.5∠BEG=45°.http://www.mofangge.com/html/qdetail/02/c2/201207/lkh3c202204495.html