如图,已知在△ABC中,AB=AC,∠B=30°,AC的垂直平分线EF交AC于点E,交BC于点F,求证CF=1/2BF
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![如图,已知在△ABC中,AB=AC,∠B=30°,AC的垂直平分线EF交AC于点E,交BC于点F,求证CF=1/2BF](/uploads/image/z/8582533-61-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2C%E2%88%A0B%3D30%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%81CF%3D1%2F2BF)
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