证明:
∵∠ACB=90°,AC=BC=1
∴∠CAB=∠CBA=45
∵△ACB绕点A旋转至△A1CB1
∴∠A1=∠CAB,∠CB1A1=∠CBA,∠A1CB1=∠ACB,B1C=BC
∴∠A1=∠B,∠CAB=∠CB1A1,B1C=AC
∵∠A1CA=∠A1CB1-∠ACD,∠BCD=∠ACB-∠ACD
∴∠A1CA=∠BCD
∴△A1CF≌△BCD (ASA)
∴CF=CD
∵AF=AC-CF,B1D=B1C-CD
∴AF=B1D
∵∠AEF=∠B1D
∴△AEF≌△B1ED (AAS)
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