题目如图,点DDD在△ABC\triangle ABC△ABC的ACACAC边上,以ADADAD为直径作△ABD\triangle ABD△ABD的外接圆,记为⊙O\odot O⊙O,∠BAD=∠CBD\angle BAD=\angle CBD∠BAD=∠CBD.(1)(1)(1)若⊙O\odot O⊙O的半径为111111,AB=17AB=17AB=17,求cos∠CBD\cos \angle CBDcos∠CBD的值;(2)(2)(2)求证:BCBCBC是⊙O\odot O⊙O的切线;(3)(3)(3)已知BFBFBF平分∠ABD\angle ABD∠ABD,交ADADAD于点EEE,交⊙O\odot O⊙O于点FFF.若tan∠BAD=155tan∠BAD=\frac{\sqrt{15}}{5}tan∠BAD=515,CD=3ODCD=3ODCD=3OD,BC=53BC=5\sqrt{3}BC=53,求BE⋅BFBE\cdot BFBE⋅BF的值.知识点:圆周角定理I章节:图形的性质 / 圆