题目如图,△ACB\triangle ACB△ACB和△ECD\triangle ECD△ECD都是等腰直角三角形,CA=CBCA=CBCA=CB,CE=CDCE=CDCE=CD,点AAA在EDEDED上.若AE=1AE=1AE=1,AD=3AD=3AD=3,则BCBCBC的长为( )A.5\sqrt{5}5B.6\sqrt{6}6C.2.52.52.5D.8\sqrt{8}8知识点:勾股定理章节:图形的性质 / 三角形