已知AB=AC,DF平行于AB.求证:①DF=DC二,②若DE⊥FC于E,那么EF=CE急························
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![已知AB=AC,DF平行于AB.求证:①DF=DC二,②若DE⊥FC于E,那么EF=CE急························](/uploads/image/z/5109226-34-6.jpg?t=%E5%B7%B2%E7%9F%A5AB%3DAC%2CDF%E5%B9%B3%E8%A1%8C%E4%BA%8EAB.%E6%B1%82%E8%AF%81%EF%BC%9A%E2%91%A0DF%3DDC%E4%BA%8C%2C%E2%91%A1%E8%8B%A5DE%E2%8A%A5FC%E4%BA%8EE%2C%E9%82%A3%E4%B9%88EF%3DCE%E6%80%A5%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7)
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已知AB=AC,DF平行于AB.求证:①DF=DC二,②若DE⊥FC于E,那么EF=CE急························
已知AB=AC,DF平行于AB.求证:①DF=DC二,②若DE⊥FC于E,那么EF=CE
急························
已知AB=AC,DF平行于AB.求证:①DF=DC二,②若DE⊥FC于E,那么EF=CE急························
证明:①∵AB=AC
∴∠B=∠C
∵DF∥AB
∴∠B=∠DFC
∵∠B=∠C
∴∠DFC=∠C
∴DF=DC
②∵AE⊥FC于E
∴∠DEC=∠DEF=90°
∵∠DFC=∠C,DE=DE
∴△DEF≌△DEC
∴EF=CE
证明②若DE⊥FC于E,那么EF=CE
连接FC,CD,做DE⊥FC,因为DF平行AB,所以角F等于角ABC,又因为AB=AC,所以又等于角C,所以角F等于角C,所以DF=DC,又因为DE垂直CF,所以EF=CE
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