(1)证明:连接OD,OF.∵BF切⊙O于点F,∴∠OFB=90°,∵弦DF⊥AB于E,且AB经过圆心O,∴DE=EF,∴BD=BF.∴∠1=∠BFD.∵OD=OF,∴∠3=∠4,∴∠ODB=∠OFB=90°,∴BD与⊙O相切;(2)解:由(1)可知∠3=∠5,∵∠2=∠5,∴∠2=∠3.又∵∠6=2∠2,∴∠6=2∠3.∵∠6+∠3=90°,∴3∠3=90°.∴∠3=30°,∵OD=2,∴DE= 3 ,∴DF=2 3 .