题目如图,在△ABC\triangle ABC△ABC中,∠C=90∘\angle C=90^{\circ}∠C=90∘,ADADAD平分∠BAC\angle BAC∠BAC,BC=30BC=30BC=30,BD:CD=3:2BD:CD=3:2BD:CD=3:2,则点DDD到ABABAB的距离为( )A.181818B.121212C.151515D.161616知识点:角平分线的性质章节:图形的性质 / 三角形