AB=AC,BD=CF,求证:DF=EFabc为一三角形,f在ac延长线上,d在ab上一点,连接fd交bc于e点.o ,是de等于ef
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![AB=AC,BD=CF,求证:DF=EFabc为一三角形,f在ac延长线上,d在ab上一点,连接fd交bc于e点.o ,是de等于ef](/uploads/image/z/9497971-19-1.jpg?t=AB%3DAC%2CBD%3DCF%2C%E6%B1%82%E8%AF%81%EF%BC%9ADF%3DEFabc%E4%B8%BA%E4%B8%80%E4%B8%89%E8%A7%92%E5%BD%A2%2Cf%E5%9C%A8ac%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%B8%8A%2Cd%E5%9C%A8ab%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E8%BF%9E%E6%8E%A5fd%E4%BA%A4bc%E4%BA%8Ee%E7%82%B9.o+%2C%E6%98%AFde%E7%AD%89%E4%BA%8Eef)
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AB=AC,BD=CF,求证:DF=EFabc为一三角形,f在ac延长线上,d在ab上一点,连接fd交bc于e点.o ,是de等于ef
AB=AC,BD=CF,求证:DF=EF
abc为一三角形,f在ac延长线上,d在ab上一点,连接fd交bc于e点.
o ,是de等于ef
AB=AC,BD=CF,求证:DF=EFabc为一三角形,f在ac延长线上,d在ab上一点,连接fd交bc于e点.o ,是de等于ef
过D作DM//AC交BC于M
则DM=DB=CF
且∠CEF=∠DEM
∠CFE=∠ACB-∠CEF
=∠B-∠DEM
=∠DMB-∠DEM
=∠MDE
故由角角边知:△DEM与△FEC全等
故DE=EF
题目写错了吧,DF不可能等于EF的!!
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