一道几何题思路求教三角形ABC,角B=65度,角C=45度,AD垂直于BC,AF是角BAC的平分线,求证角DAF=二分之(角B减角C),思路是什么?
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![一道几何题思路求教三角形ABC,角B=65度,角C=45度,AD垂直于BC,AF是角BAC的平分线,求证角DAF=二分之(角B减角C),思路是什么?](/uploads/image/z/14698443-3-3.jpg?t=%E4%B8%80%E9%81%93%E5%87%A0%E4%BD%95%E9%A2%98%E6%80%9D%E8%B7%AF%E6%B1%82%E6%95%99%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2C%E8%A7%92B%3D65%E5%BA%A6%2C%E8%A7%92C%3D45%E5%BA%A6%2CAD%E5%9E%82%E7%9B%B4%E4%BA%8EBC%2CAF%E6%98%AF%E8%A7%92BAC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E6%B1%82%E8%AF%81%E8%A7%92DAF%3D%E4%BA%8C%E5%88%86%E4%B9%8B%EF%BC%88%E8%A7%92B%E5%87%8F%E8%A7%92C%29%2C%E6%80%9D%E8%B7%AF%E6%98%AF%E4%BB%80%E4%B9%88%3F)
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