题目如图,在四边形ABCDABCDABCD中,AB,AB,AB∥CDCDCD,AD⊥ABAD\bot ABAD⊥AB,以DDD为圆心,ADADAD为半径的弧恰好与BCBCBC相切,切点为EEE,若ABCD=13\frac{AB}{CD}=\frac{1}{3}CDAB=31,则sinC\sin CsinC的值是( )A.23\frac{2}{3}32B.53\frac{\sqrt{5}}{3}35C.34\frac{3}{4}43D.74\frac{\sqrt{7}}{4}47知识点:切线长定理章节:图形的性质 / 圆