如图,AB=AD,AC=AE,∠DAB=∠CAE,BE与DC交于点P.求证:PA平分∠DPE快点啊啊
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![如图,AB=AD,AC=AE,∠DAB=∠CAE,BE与DC交于点P.求证:PA平分∠DPE快点啊啊](/uploads/image/z/4052007-63-7.jpg?t=%E5%A6%82%E5%9B%BE%2CAB%3DAD%2CAC%3DAE%2C%E2%88%A0DAB%3D%E2%88%A0CAE%2CBE%E4%B8%8EDC%E4%BA%A4%E4%BA%8E%E7%82%B9P.%E6%B1%82%E8%AF%81%3APA%E5%B9%B3%E5%88%86%E2%88%A0DPE%E5%BF%AB%E7%82%B9%E5%95%8A%E5%95%8A)
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如图,AB=AD,AC=AE,∠DAB=∠CAE,BE与DC交于点P.求证:PA平分∠DPE快点啊啊
如图,AB=AD,AC=AE,∠DAB=∠CAE,BE与DC交于点P.求证:PA平分∠DPE
快点啊啊
如图,AB=AD,AC=AE,∠DAB=∠CAE,BE与DC交于点P.求证:PA平分∠DPE快点啊啊
(你的图画得不好,影响答题)
思路:如果PA是∠DPE的平分线,则点A到角的两边(PD和PE)的距离应相等.过A作PD的垂线交于H1,过A作PE的垂线交于H2.则AH1为△ADC的高,AH2为△ABE的高,而△ADC≌△ABE可证,得出AH1=AH2,此题可解.
∵∠DAB=∠CAE
∴∠DAC=∠BAE
∵在△ADC和△ABE中,AD=AB,AC=AE,∠DAC=∠BAE
∴△ADC≌△ABE
过A作PD的垂线交于H1,过A作PE的垂线交于H2.则AH1为△ADC的高,AH2为△ABE的高,
∴AH1=AH2,AH1⊥PD,AH2⊥PE
∴A在∠DPE的平分线上,PA平分∠DPE
∵AD=AB, AC=AE,
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