如图,已知D是△ABC中BC边上一点,E是AD上一点,EB=EC ,∠ABE=∠ACE,求证∠BAE=∠CAE.注意:用角平分线的性质
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![如图,已知D是△ABC中BC边上一点,E是AD上一点,EB=EC ,∠ABE=∠ACE,求证∠BAE=∠CAE.注意:用角平分线的性质](/uploads/image/z/8574443-35-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5D%E6%98%AF%E2%96%B3ABC%E4%B8%ADBC%E8%BE%B9%E4%B8%8A%E4%B8%80%E7%82%B9%26%238203%3B%2CE%E6%98%AFAD%E4%B8%8A%E4%B8%80%E7%82%B9%2CEB%3DEC+%2C%E2%88%A0ABE%3D%E2%88%A0ACE%2C%E6%B1%82%E8%AF%81%E2%88%A0BAE%3D%E2%88%A0CAE.%E6%B3%A8%E6%84%8F%EF%BC%9A%E7%94%A8%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E7%9A%84%E6%80%A7%E8%B4%A8)
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