题目如图,在平行四边形ABCDABCDABCD中,CD=2ADCD=2ADCD=2AD,BE⊥ADBE\bot ADBE⊥AD于点EEE,FFF为DCDCDC的中点,连接EFEFEF,BFBFBF.下列结论中正确的是( )A.EF=BFEF=BFEF=BFB.∠ABE=∠EBF\angle ABE=\angle EBF∠ABE=∠EBFC.S△CFB=S△EFBS_{\triangle CFB}=S_{\triangle EFB}S△CFB=S△EFBD.△CBF\triangle CBF△CBF是等边三角形知识点:平行四边形的性质章节:未标注