在三角形ABC中,BD是角ABC的平分线,在三角形ABC外一点E,使得角EAB=角ACB.AE=DC,ED与AB交k,求证kE=kD
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![在三角形ABC中,BD是角ABC的平分线,在三角形ABC外一点E,使得角EAB=角ACB.AE=DC,ED与AB交k,求证kE=kD](/uploads/image/z/12390417-9-7.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CBD%E6%98%AF%E8%A7%92ABC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E5%A4%96%E4%B8%80%E7%82%B9E%2C%E4%BD%BF%E5%BE%97%E8%A7%92EAB%3D%E8%A7%92ACB.AE%3DDC%2CED%E4%B8%8EAB%E4%BA%A4k%2C%E6%B1%82%E8%AF%81kE%3DkD)
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