证明：连接AC.
∵AB=BC,∴∠BAC=∠BCA,
∵AB∥CD,∴∠BAC=∠DCA,
∴∠DCA=∠BCA,
又∠D=∠AEC=90°,AC=AC,
∴△ACD≅△ACE,
∴AD=AE
  
∵AD⊥DC,AB∥CD,
∴AD⊥AB,
∴直线AD与圆O相切