题目如图,PAPAPA为⊙O\odot O⊙O的切线,AAA为切点,过点AAA作AB⊥OPAB\bot OPAB⊥OP,垂足为点CCC,交⊙O\odot O⊙O于点BBB,延长BOBOBO与PAPAPA的延长线交于点DDD.(1)(1)(1)求证:PBPBPB为⊙O\odot O⊙O的切线;(2)(2)(2)若OB=3OB=3OB=3,OD=5OD=5OD=5,求OPOPOP的长.知识点:切线的判定与性质章节:图形的性质 / 圆