如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.
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![如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.](/uploads/image/z/1816948-28-8.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E5%B7%B2%E7%9F%A5%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2C%E2%88%A0BAC%3D120%C2%B0%2CAC%E7%9A%84%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BFEF%E4%BA%A4AC%E4%BA%8E%E7%82%B9E%2C%E4%BA%A4BC%E4%BA%8E%E7%82%B9F%2C%E6%B1%82%E8%AF%81%EF%BC%9ABF%3D2CF.%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E5%B7%B2%E7%9F%A5%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2C%E2%88%A0BAC%3D120%C2%B0%2CAC%E7%9A%84%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BFEF%E4%BA%A4AC%E4%BA%8E%E7%82%B9E%2C%E4%BA%A4BC%E4%BA%8E%E7%82%B9F%2C%E6%B1%82%E8%AF%81%EF%BC%9ABF%3D2CF.)
如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.
如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.
如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,
求证:BF=2CF.
如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.如图所示,已知在△ABC中,AB=AC,∠BAC=120°,AC的垂直平分线EF交AC于点E,交BC于点F,求证:BF=2CF.
连接AF,因为EF是AC的垂直平分线,∴∠FAC=∠C=∠B=30°,AF=CF
∴∠BAF=∠BAC-∠FAC=90°,30度角所对的直角边等于斜边的一半,∴ BF=2AF=2CF
连接AF
∵AB=AC,∠BAC=120°
∴∠B=∠C=30°
∵EF垂直平分AC
∴AF=CF
∴∠EAF=∠C=30°
∴∠BAF=∠BAC-∠EAF=120°-30°=90°
∴在Rt△ABF中,∠B=30°
∴AF=1/2BF
∵AF=CF
∴CF=1/2BF
即BF=2CF
∵△ABC是等腰三角形,∠BAC=120°
∴∠B=∠C=30°
连接AF,
∵EF是AC的c垂直平分线
∴AF=FC ∠FAC=∠C=30°
∴∠BAF=∠BAC-∠FAC=120°-30°=90°
在△BAF中
∵ ∠BAF=90° ∠B=30°
∴BF=2AF=2FC