题目如图,在矩形ABCDABCDABCD中,AB=2AB=2AB=2,BC=221BC=2\sqrt{21}BC=221,EEE是BCBCBC的中点,将△ABE\triangle ABE△ABE沿直线AEAEAE翻折,点BBB落在点FFF处,连接CFCFCF,则CFCFCF的长为( )A.888B.425\frac{42}{5}542C.172\frac{17}{2}217D.9215\frac{9\sqrt{21}}{5}5921知识点:轴对称变换章节:图形的变化 / 图形的旋转