题目如图,在平行四边形ABCDABCDABCD中,点EEE在边ADADAD上,BABABA,CECECE的延长线交于点FFF.若AF=1AF=1AF=1,AB=2AB=2AB=2,则AEAD=( )\frac{AE}{AD}=\left(\ \ \right)ADAE=( )A.12\frac{1}{2}21B.13\frac{1}{3}31C.14\frac{1}{4}41D.15\frac{1}{5}51知识点:平行四边形的性质章节:未标注