已知三角形ABC和三角形DCB均是等边三角形,点B、C、E在同一条直线上,AE与BD交于点O,AE与CD交于点G,AC与BD交于点F,连接OC、FG,则下列结论:1、AE=BD;2、AG=BF;3、FG平行BE;4、角BOC=角EOC,其中正确的
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![已知三角形ABC和三角形DCB均是等边三角形,点B、C、E在同一条直线上,AE与BD交于点O,AE与CD交于点G,AC与BD交于点F,连接OC、FG,则下列结论:1、AE=BD;2、AG=BF;3、FG平行BE;4、角BOC=角EOC,其中正确的](/uploads/image/z/6937064-8-4.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E5%92%8C%E4%B8%89%E8%A7%92%E5%BD%A2DCB%E5%9D%87%E6%98%AF%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E7%82%B9B%E3%80%81C%E3%80%81E%E5%9C%A8%E5%90%8C%E4%B8%80%E6%9D%A1%E7%9B%B4%E7%BA%BF%E4%B8%8A%2CAE%E4%B8%8EBD%E4%BA%A4%E4%BA%8E%E7%82%B9O%2CAE%E4%B8%8ECD%E4%BA%A4%E4%BA%8E%E7%82%B9G%2CAC%E4%B8%8EBD%E4%BA%A4%E4%BA%8E%E7%82%B9F%2C%E8%BF%9E%E6%8E%A5OC%E3%80%81FG%2C%E5%88%99%E4%B8%8B%E5%88%97%E7%BB%93%E8%AE%BA%EF%BC%9A1%E3%80%81AE%3DBD%EF%BC%9B2%E3%80%81AG%3DBF%EF%BC%9B3%E3%80%81FG%E5%B9%B3%E8%A1%8CBE%EF%BC%9B4%E3%80%81%E8%A7%92BOC%3D%E8%A7%92EOC%2C%E5%85%B6%E4%B8%AD%E6%AD%A3%E7%A1%AE%E7%9A%84)
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