题目如图,△ABC\triangle ABC△ABC是⊙O\odot O⊙O的内接三角形,ABABAB为⊙O\odot O⊙O的直径,CDCDCD平分∠ACB\angle ACB∠ACB,交⊙O\odot O⊙O于点DDD,连接ADADAD,点EEE在弦CDCDCD上,且ED=ADED=ADED=AD,连接AEAEAE.(1)(1)(1)求证:∠BAE=∠CAE\angle BAE=\angle CAE∠BAE=∠CAE;(2)(2)(2)若∠B=60∘\angle B=60^{\circ}∠B=60∘,AB=8AB=8AB=8,求AEAEAE的长.知识点:垂径定理章节:图形的性质 / 圆