题目如图,已知Rt△ABCRt\triangle ABCRt△ABC内接于⊙O\odot O⊙O,∠C=90∘\angle C=90^{\circ}∠C=90∘.点DDD是BC^\widehat {BC}BC的中点,BEBEBE平分∠ABC\angle ABC∠ABC,交ADADAD于点EEE,连结BDBDBD、CDCDCD.(1)(1)(1)若∠CAD=α\angle CAD=\alpha∠CAD=α,请用α\alphaα表示∠ABE\angle ABE∠ABE;(2)(2)(2)证明:AD=AE+CDAD=AE+CDAD=AE+CD;(3)(3)(3)若AC=6AC=6AC=6,AB=10AB=10AB=10,求BDBDBD的长.知识点:切线的性质章节:图形的性质 / 圆