如图,AB=AC,AD = AE,CD=BE.求证:∠DAB=∠EAC.
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如图,AB=AC,AD=AE,CD=BE.求证:∠DAB=∠EAC.如图,AB=AC,AD=AE,CD=BE.求证:∠DAB=∠EAC.如图,AB=AC,AD=AE,CD=BE.求证:∠DAB=∠EA
如图,AB=AC,AD = AE,CD=BE.求证:∠DAB=∠EAC.
如图,AB=AC,AD = AE,CD=BE.求证:∠DAB=∠EAC.
如图,AB=AC,AD = AE,CD=BE.求证:∠DAB=∠EAC.
∵AB=AC,AD = AE,CD=BE
∴△ADC≌△AEB(SSS)
∴∠DAC=∠EAB
∴∠DAC-∠BAC=∠EAB-∠BAC
∴∠DAB=∠EAC
因为AB=AC,AD = AE,CD=BE
三角形adc全等于aeb
∵AB=AC,AD = AE,CD=BE
∴△ADC≌△AEB
∴∠DAC=∠BAE
∴∠BAD=∠CAE
AB=AC
所以∠E=∠D
因为AD=AE
AC=AB
所以三角形ADC全等于三角形AEB
∠DAB+∠BAC=∠EAC+∠BAC
即∠DAC=∠EAB
所以∠DAB=∠EAC
边边边证明三角形全等
得到∠DAC=∠EAB
减去公共角BAC
可证∠DAB=∠EAC
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