题目如图,将△ABC\triangle ABC△ABC绕点CCC顺时针旋转,点BBB的对应点为点EEE,点AAA的对应点为点DDD,当点EEE恰好落在边ACACAC上时,连接ADADAD,∠ACB=36∘\angle ACB=36^{\circ}∠ACB=36∘,AB=BCAB=BCAB=BC,AC=2AC=2AC=2,则ABABAB的长度是( )A.5−1\sqrt{5}-15−1B.111C.5−12\frac{\sqrt{5}-1}{2}25−1D.32\frac{3}{2}23知识点:旋转的性质章节:未标注