已知,如图所示,AB=AC,AD=AE,∠DAB=∠CAE,CD与BE相交于P、求证 PA平分∠DPE图在这
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![已知,如图所示,AB=AC,AD=AE,∠DAB=∠CAE,CD与BE相交于P、求证 PA平分∠DPE图在这](/uploads/image/z/5563781-53-1.jpg?t=%E5%B7%B2%E7%9F%A5%2C%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2CAB%3DAC%2CAD%3DAE%2C%E2%88%A0DAB%3D%E2%88%A0CAE%2CCD%E4%B8%8EBE%E7%9B%B8%E4%BA%A4%E4%BA%8EP%E3%80%81%E6%B1%82%E8%AF%81+PA%E5%B9%B3%E5%88%86%E2%88%A0DPE%E5%9B%BE%E5%9C%A8%E8%BF%99)
已知,如图所示,AB=AC,AD=AE,∠DAB=∠CAE,CD与BE相交于P、求证 PA平分∠DPE图在这
已知,如图所示,AB=AC,AD=AE,∠DAB=∠CAE,CD与BE相交于P、求证 PA平分∠DPE
图在这
已知,如图所示,AB=AC,AD=AE,∠DAB=∠CAE,CD与BE相交于P、求证 PA平分∠DPE图在这
做辅助线BD和CE,利用SAS定理得:BE=CD,BD=CE,又BC=BC,则三角形BCD全等于三角形CBE,则,∠CBE=∠BCD,则,BP=CP又AB=AC,AP=AP,就可证
如图所示,已知AB⊥AC,AD⊥AE,AB=AC,AD=AE,求证:BE=DC
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如图所示,已知AB=AC,AD=AE,请说明BD=CE
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