如图:把△ABE绕点A逆时针旋转90度,得到△ADG,则△ABE≌△ADG,∠EAG=∠BAD=90°, ∴∠ABE=∠ADG=90°,AE=AG,BE=DG, ∴∠FDG=∠FDA+∠ADG=90°+90°=180°, ∴F、D、G三点共线. ∵EF=BE+DF, ∴EF=DG+DF=GF. ∵在△AGF与△AEF中, , ∴△AGF≌△AEF(SSS), ∴∠GAF=∠EAF,∠1=∠2, ∵∠GAF+∠EAF=∠EAG=90°, ∴∠EAF=×90°=45°,故③正确; ∵∠1=∠2,AD⊥FG于D,AH⊥EF于H, ∴AD=AH, ∵AD=AB, ∴AH=AB, 又∵AH⊥EF于H,AB⊥BC于B, ∴AE平分∠BEF,故①正确; ∵AE平分∠BEF, ∴∠AEB=∠AEH, ∵∠AEB+∠BAE=90°,∠AEH+∠HAE=90°, ∴∠BAE=∠HAE, 又∵EH⊥AH于H,EB⊥AB于B, ∴BE=HE, ∵BE=DG, ∴HE=DG, ∵EF=HE+FH,GF=DG+FD,EF=GF, ∴FH=FD,故②正确; ∵△AEF≌△AGF, ∴S△EAF=S△GAF. ∵△ABE≌△ADG, ∴S△GAF=S△ADG+S△ADFS△ABE+S△ADF, ∴S△EAF=S△ABE+S△ADF,故④正确; ∵EF=HE+FH,BE=HE,FH=FD, ∴EF=BE+FD, ∴△CEF的周长=EF+EC+CF=BE+FD+EC+CF=BC+CD=2AB=2,故⑤正确. 故选D.
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