题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,AB=AC=6AB=AC=6AB=AC=6,AE=2AE=2AE=2,过点CCC作CDCDCD垂直于BEBEBE延长线于点DDD,连接ADADAD,则ADADAD的长为( )A.222\sqrt{2}22B.323\sqrt{2}32C.2105\frac{{2\sqrt{10}}}{5}5210D.655\frac{{6\sqrt{5}}}{5}565知识点:勾股定理章节:图形的性质 / 三角形