摘要: 猜想结果:图2结论S△PBC=S△PAC+S△PCD, 图3结论S△PBC=S△PAC-S△PCD 证明:如图2.过点P作EF垂直AD.分别交AD.BC于E.F两点. ∵ S△PBC=BC·PF=BC·PE+BC·EF =AD·PE+BC·EF=S△PAD+S矩形ABCD S△PAC+S△PCD=S△PAD+S△ADC=S△PAD+S矩形ABCD ∴ S△PBC=S△PAC+S△PCD

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