在三角形ABC中,AD是BC边的中线,DE平分角ADB交AB边于E点,DF平分角ADC交AC于F点,求证:BE+CF>EF.用三角形的全等和角平分线证明.
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![在三角形ABC中,AD是BC边的中线,DE平分角ADB交AB边于E点,DF平分角ADC交AC于F点,求证:BE+CF>EF.用三角形的全等和角平分线证明.](/uploads/image/z/3001534-70-4.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CAD%E6%98%AFBC%E8%BE%B9%E7%9A%84%E4%B8%AD%E7%BA%BF%2CDE%E5%B9%B3%E5%88%86%E8%A7%92ADB%E4%BA%A4AB%E8%BE%B9%E4%BA%8EE%E7%82%B9%2CDF%E5%B9%B3%E5%88%86%E8%A7%92ADC%E4%BA%A4AC%E4%BA%8EF%E7%82%B9%2C%E6%B1%82%E8%AF%81%EF%BC%9ABE%2BCF%3EEF.%E7%94%A8%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%85%A8%E7%AD%89%E5%92%8C%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E8%AF%81%E6%98%8E.)
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