已知:如图AB=AC,AD=AE.AB,DC相交于点M,AC,BE相交于点N,
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已知:如图AB=AC,AD=AE.AB,DC相交于点M,AC,BE相交于点N,
![已知:如图AB=AC,AD=AE.AB,DC相交于点M,AC,BE相交于点N,](/uploads/image/z/4046227-43-7.jpg?t=%E5%B7%B2%E7%9F%A5%3A%E5%A6%82%E5%9B%BEAB%3DAC%2CAD%3DAE.AB%2CDC%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9M%2CAC%2CBE%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9N%2C)
角DAB=角EAC,所以角DAC=角EAB,所以三角形ADC全等于 三角形AEB,所以 AM=AN;
OVER
答案补充角DAB=角EAC,所以角DAC=角EAB(边角边(SAS))
答案补充角DAB=角EAC,所以角DAC=角EAB,AB=AC,AD=AE (边角边
OVER
答案补充角DAB=角EAC,所以角DAC=角EAB(边角边(SAS))
答案补充角DAB=角EAC,所以角DAC=角EAB,AB=AC,AD=AE (边角边
已知:如图AB=AC,AD=AE.AB,DC相交于点M,AC,BE相交于点N,
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真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an
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