AE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BEAE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BE
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![AE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BEAE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BE](/uploads/image/z/2697977-65-7.jpg?t=AE%E4%BA%A4BC%E4%BA%8ED%2C%E8%A7%921%3D%E8%A7%922%3D%E8%A7%923%2CAB%3DAD+%E6%B1%82%E8%A7%92ADC%3D%E8%A7%92ABE+DC%3DBEAE%E4%BA%A4BC%E4%BA%8ED%2C%E8%A7%921%3D%E8%A7%922%3D%E8%A7%923%2CAB%3DAD+%E6%B1%82%E8%A7%92ADC%3D%E8%A7%92ABE+DC%3DBE)
AE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BEAE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BE
AE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BE
AE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BE
AE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BEAE交BC于D,角1=角2=角3,AB=AD 求角ADC=角ABE DC=BE
是求证吧?
角ADC=角2+角ABC(三角形外角)
角ABE=角3+角ABC
角3=角2
所以角ADC=角ABE
因为角1=角2,AB=A,角ADC=角ABE
所以三角形ADC和三角形ABE全等(角边角)
所以DC=BE