题目如图,在△ABC\triangle ABC△ABC中,BDBDBD是中线.(1)(1)(1)如图(1),延长BDBDBD至点EEE,使得DE=BDDE=BDDE=BD,连接AEAEAE.①求证:△ADE\triangle ADE△ADE≌△CDB\triangle CDB△CDB;②若AB=6AB=6AB=6,BC=4BC=4BC=4,设BD=xBD=xBD=x,直接写出xxx的取值范围;(2)(2)(2)如图(2),延长CACACA到点FFF,使AF=BCAF=BCAF=BC,若∠ABC=∠BAC\angle ABC=\angle BAC∠ABC=∠BAC,求证:BF=2BDBF=2BDBF=2BD.知识点:切线的判定章节:图形的性质 / 圆