如图,在等腰△ABC中,CH是底边是的高线,点P是线段CH上不与端点重合的任意一点,连接AP并延长交BC于点E,连接BP并延长交AC与点P (1)证明:角CAP=角CBP (2)证明:AE=BF (3)以线段AE,BF和AB为边构成一个新的
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![如图,在等腰△ABC中,CH是底边是的高线,点P是线段CH上不与端点重合的任意一点,连接AP并延长交BC于点E,连接BP并延长交AC与点P (1)证明:角CAP=角CBP (2)证明:AE=BF (3)以线段AE,BF和AB为边构成一个新的](/uploads/image/z/625077-45-7.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%AD%89%E8%85%B0%E2%96%B3ABC%E4%B8%AD%2CCH%E6%98%AF%E5%BA%95%E8%BE%B9%E6%98%AF%E7%9A%84%E9%AB%98%E7%BA%BF%2C%E7%82%B9P%E6%98%AF%E7%BA%BF%E6%AE%B5CH%E4%B8%8A%E4%B8%8D%E4%B8%8E%E7%AB%AF%E7%82%B9%E9%87%8D%E5%90%88%E7%9A%84%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2C%E8%BF%9E%E6%8E%A5AP%E5%B9%B6%E5%BB%B6%E9%95%BF%E4%BA%A4BC%E4%BA%8E%E7%82%B9E%2C%E8%BF%9E%E6%8E%A5BP%E5%B9%B6%E5%BB%B6%E9%95%BF%E4%BA%A4AC%E4%B8%8E%E7%82%B9P+%281%29%E8%AF%81%E6%98%8E%3A%E8%A7%92CAP%3D%E8%A7%92CBP+%282%29%E8%AF%81%E6%98%8E%3AAE%3DBF+%283%29%E4%BB%A5%E7%BA%BF%E6%AE%B5AE%2CBF%E5%92%8CAB%E4%B8%BA%E8%BE%B9%E6%9E%84%E6%88%90%E4%B8%80%E4%B8%AA%E6%96%B0%E7%9A%84)
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