.已知矩形ABCD和点P,当点P在图①中的位置时,则有结论:S△PBC=S△PAC+S△PCD.理由:过点P作EF⊥BC,分别交AD、BC于E、F两点.∵,S△PBC+S△PAD=1/2BC X PF+1/2AD X PE=1/2BC(PF+PE)=1/2BC X EF=1/2S矩形ABCD.又∵,S△
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![.已知矩形ABCD和点P,当点P在图①中的位置时,则有结论:S△PBC=S△PAC+S△PCD.理由:过点P作EF⊥BC,分别交AD、BC于E、F两点.∵,S△PBC+S△PAD=1/2BC X PF+1/2AD X PE=1/2BC(PF+PE)=1/2BC X EF=1/2S矩形ABCD.又∵,S△](/uploads/image/z/8402644-28-4.jpg?t=.%E5%B7%B2%E7%9F%A5%E7%9F%A9%E5%BD%A2ABCD%E5%92%8C%E7%82%B9P%2C%E5%BD%93%E7%82%B9P%E5%9C%A8%E5%9B%BE%E2%91%A0%E4%B8%AD%E7%9A%84%E4%BD%8D%E7%BD%AE%E6%97%B6%2C%E5%88%99%E6%9C%89%E7%BB%93%E8%AE%BA%EF%BC%9AS%E2%96%B3PBC%EF%BC%9DS%E2%96%B3PAC%EF%BC%8BS%E2%96%B3PCD.%E7%90%86%E7%94%B1%EF%BC%9A%E8%BF%87%E7%82%B9P%E4%BD%9CEF%E2%8A%A5BC%2C%E5%88%86%E5%88%AB%E4%BA%A4AD%E3%80%81BC%E4%BA%8EE%E3%80%81F%E4%B8%A4%E7%82%B9.%E2%88%B5%2CS%E2%96%B3PBC%2BS%E2%96%B3PAD%3D1%2F2BC+X+PF%2B1%2F2AD+X+PE%3D1%2F2BC%28PF%2BPE%29%3D1%2F2BC+X+EF%3D1%2F2S%E7%9F%A9%E5%BD%A2ABCD.%E5%8F%88%E2%88%B5%2CS%E2%96%B3)
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