20.解:(1)连接EB,务:B∵在梯形ABCD中,∠BAD=∠ADC=90°,CD=2,AD=AB=1,BD=√2,BC∴BD2 BC2=CD2,∴BC⊥BD平面BDEF⊥平面ABCD,平面BDEF∩平面ABCD=BD,BCC平面ABCD,∴BC⊥平面BDEF,∴BC⊥DF∵在正方形BDEF中,DF⊥EB,EB∩BC=B∴DF⊥平面BCE∵CEC平面BCE,∴DF⊥CE(6分)(2)在AE上存在点G,当AG=1时,使得平面OBGGE 2∥平面EFC证明如下:∵AB∥DC,AB=1,DC=2,Ao 1又GE=2…OG∥CE,又OG≠平面EFC,CEC平面EFC,∴OG∥平面EFCEF∥OB,EFC平面EFC,OB¢平面EFC,∴OB∥平面EFC∵OB∩OG=0,∴∵平面OBG∥平面EFC.(12分)
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