已知,延长△ABC的各边,使得BF=AC,AE=CD=AB,顺次连接点D.E.F,得到△DEF为等边三角形.求证:△AEF全等于△CDE全等于△BFD
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![已知,延长△ABC的各边,使得BF=AC,AE=CD=AB,顺次连接点D.E.F,得到△DEF为等边三角形.求证:△AEF全等于△CDE全等于△BFD](/uploads/image/z/2557894-22-4.jpg?t=%E5%B7%B2%E7%9F%A5%2C%E5%BB%B6%E9%95%BF%E2%96%B3ABC%E7%9A%84%E5%90%84%E8%BE%B9%2C%E4%BD%BF%E5%BE%97BF%3DAC%2CAE%3DCD%3DAB%2C%E9%A1%BA%E6%AC%A1%E8%BF%9E%E6%8E%A5%E7%82%B9D.E.F%2C%E5%BE%97%E5%88%B0%E2%96%B3DEF%E4%B8%BA%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2.%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3AEF%E5%85%A8%E7%AD%89%E4%BA%8E%E2%96%B3CDE%E5%85%A8%E7%AD%89%E4%BA%8E%E2%96%B3BFD)
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