题目如图,△ABC\triangle ABC△ABC内接于⊙O\odot O⊙O,DEDEDE,FGFGFG是⊙O\odot O⊙O的弦,AB=DEAB=DEAB=DE,FG=ACFG=ACFG=AC.下列结论:①DE+FG=BCDE+FG=BCDE+FG=BC;②DE^+FG^=BC^\widehat {DE}+\widehat {FG}=\widehat {BC}DE+FG=BC;③∠DOE+∠FOG=∠BOC\angle DOE+\angle FOG=\angle BOC∠DOE+∠FOG=∠BOC;④∠DEO+∠EGO=∠BAC\angle DEO+\angle EGO=\angle BAC∠DEO+∠EGO=∠BAC,其中所有正确结论的序号是( )A.①②③④B.②③C.②④D.②③④知识点:切线的判定章节:图形的性质 / 圆