题目如图,矩形ABCDABCDABCD,AB=6AB=6AB=6,BC=3BC=3BC=3,点EEE是边ABABAB上的动点,点FFF是射线BCBCBC上的动点,且BF=2AEBF=2AEBF=2AE,连接AFAFAF,CECECE.若12AF+CE=m\frac{1}{2}AF+CE=m21AF+CE=m,则mmm的最小值为( )A.323\sqrt{2}32B.353\sqrt{5}35C.626\sqrt{2}62D.656\sqrt{5}65知识点:矩形的性质章节:未标注