题目如图,ABABAB为⊙O\odot O⊙O的直径,点FFF在⊙O\odot O⊙O上,OF⊥ABOF\bot ABOF⊥AB,点PPP在ABABAB的延长线上,PCPCPC与⊙O\odot O⊙O相切于点CCC,与OFOFOF的延长线相交于点DDD,ACACAC与OFOFOF相交于点EEE.(1)(1)(1)求证:DC=DEDC=DEDC=DE;(2)(2)(2)若0A=2OE0A=2OE0A=2OE,DF=3DF=3DF=3,求PBPBPB的长.知识点:切线的判定章节:图形的性质 / 圆