题目已知:BCBCBC是⊙O\odot O⊙O的直径,AAA是⊙O\odot O⊙O上一点,AD⊥BCAD\bot BCAD⊥BC,垂足为DDD,AB^=AE^\widehat {AB}=\widehat {AE}AB=AE,BEBEBE交ADADAD的延长线于点FFF,延长BEBEBE、ACACAC交于点GGG.求证:BF=FGBF=FGBF=FG.知识点：垂径定理章节：图形的性质 / 圆
已知:是的直径,是上一点,,垂足为,,交的延长线于点,延长、交于点.求证:.已知:是的直径,是上一点,,垂足为,,交的延长线于点,延长、交于点.求证:.已知:是的直径,是上一点,,垂足为,,交的延长线于点,延长、交于点.求证:.