题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,∠C=90∘\angle C=90^{\circ}∠C=90∘,∠BAC\angle BAC∠BAC的平分线ADADAD交BCBCBC于点DDD,CD=3CD=3CD=3,则点DDD到ABABAB的距离是( )A.666B.222C.333D.444知识点:点到直线的距离章节:未标注