题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,∠ACB=90∘\angle ACB=90^{\circ}∠ACB=90∘,CD⊥ABCD\bot ABCD⊥AB,垂足为DDD,AFAFAF平分∠CAB\angle CAB∠CAB,交CDCDCD于点EEE,交CBCBCB于点FFF.若AC=3AC=3AC=3,AB=5AB=5AB=5,则CECECE的长为( )A.32\frac{3}{2}23B.43\frac{4}{3}34C.53\frac{5}{3}35D.85\frac{8}{5}58知识点:相似三角形的判定I章节:图形的变化 / 图形的相似