题目如图,在矩形ABCDABCDABCD中,点EEE在边ADADAD上,△CDE\triangle CDE△CDE沿CECECE折叠得到△CFE\triangle CFE△CFE,且点BBB,FFF,EEE三点共线,连接DFDFDF,若BE=256BE=\frac{25}{6}BE=625,DE=3DE=3DE=3,则AE=AE=AE=______,DF=______.DF=\_\_\_\_\_\_.DF=______.知识点:轴对称变换章节:未标注