△ABC中,AB=AC,AD⊥BC,垂足为点D,∠BAC=48°,CE、CF三等分∠ACB,分别交AD于点E、F,连接BE并延长,交AC于点G,连接FG,则∠AGF=?FG和CE不是平行的,画的可能能有点像
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![△ABC中,AB=AC,AD⊥BC,垂足为点D,∠BAC=48°,CE、CF三等分∠ACB,分别交AD于点E、F,连接BE并延长,交AC于点G,连接FG,则∠AGF=?FG和CE不是平行的,画的可能能有点像](/uploads/image/z/10156569-33-9.jpg?t=%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2CAD%E2%8A%A5BC%2C%E5%9E%82%E8%B6%B3%E4%B8%BA%E7%82%B9D%2C%E2%88%A0BAC%3D48%C2%B0%2CCE%E3%80%81CF%E4%B8%89%E7%AD%89%E5%88%86%E2%88%A0ACB%2C%E5%88%86%E5%88%AB%E4%BA%A4AD%E4%BA%8E%E7%82%B9E%E3%80%81F%2C%E8%BF%9E%E6%8E%A5BE%E5%B9%B6%E5%BB%B6%E9%95%BF%2C%E4%BA%A4AC%E4%BA%8E%E7%82%B9G%2C%E8%BF%9E%E6%8E%A5FG%2C%E5%88%99%E2%88%A0AGF%3D%3FFG%E5%92%8CCE%E4%B8%8D%E6%98%AF%E5%B9%B3%E8%A1%8C%E7%9A%84%2C%E7%94%BB%E7%9A%84%E5%8F%AF%E8%83%BD%E8%83%BD%E6%9C%89%E7%82%B9%E5%83%8F)
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