题目如图,在△ABC\triangle ABC△ABC中,∠ACB=90∘\angle ACB=90^{\circ}∠ACB=90∘,点OOO为BCBCBC边上一点,以点OOO为圆心,OBOBOB长为半径的圆与边ABABAB相交于点DDD,连接DCDCDC,且DC=ACDC=ACDC=AC.(1)(1)(1)求证:DCDCDC为⊙O\odot O⊙O的切线;(2)(2)(2)若⊙O\odot O⊙O的半径为333,CD=4CD=4CD=4,求BCBCBC的长.知识点:切线的判定与性质章节:图形的性质 / 圆