①∵∠EAB+∠BAP=90°,∠PAD+∠BAP=90°, ∴∠EAB=∠PAD, 又∵AE=AP,AB=AD, ∴△APD≌△AEB(故①正确); ③∵△APD≌△AEB, ∴∠APD=∠AEB, 又∵∠AEB=∠AEP+∠BEP,∠APD=∠AEP+∠PAE, ∴∠BEP=∠PAE=90°, ∴EB⊥ED(故③正确); ②过B作BF⊥AE,交AE的延长线于F, ∵AE=AP,∠EAP=90°, ∴∠AEP=∠APE=45°, 又∵③中EB⊥ED,BF⊥AF, ∴∠FEB=∠FBE=45°, 又∵BE===, ∴BF=EF=(故②不正确); ④如图,连接BD,在Rt△AEP中, ∵AE=AP=1, ∴EP=, 又∵PB=, ∴BE=, ∵△APD≌△AEB, ∴PD=BE=, ∴S△ABP+S△ADP=S△ABD-S△BDP=S正方形ABCD-×DP×BE=×(4+)-××=+.(故④不正确). ⑤∵EF=BF=,AE=1, ∴在Rt△ABF中,AB2=(AE+EF)2+BF2=4+, ∴S正方形ABCD=AB2=4+(故⑤正确); 故选:D. |