如图,∠BAC=∠DAE,∠ABD=∠ACE,BD=CE.求证:AB=AC
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如图,∠BAC=∠DAE,∠ABD=∠ACE,BD=CE.求证:AB=AC
如图,∠BAC=∠DAE,∠ABD=∠ACE,BD=CE.求证:AB=AC
如图,∠BAC=∠DAE,∠ABD=∠ACE,BD=CE.求证:AB=AC
∵∠BAC=∠DAE
∴∠BAD+∠DAC=∠DAC+∠CAE
∴∠BAD=∠CAE
又∵∠ABD=∠ACE,BD=CE
∴△ABD≌△ACE(AAS)
∴AB=AC
设DE交AC与F
因为∠BAC=∠DAE
所以∠BAD+∠DAF=∠DAF+∠FAE
所以∠BAD=∠FAE
在△ADB和△AEC中
因为∠BAD=∠FAE
∠ABD=∠ACE
BD=CE
所以△ADB≌△AEC
所以AB=AC(明白不)
设DE交AC与F
因为∠BAC=∠DAE
所以∠BAD+∠DAF=∠DAF+∠FAE
所以∠BAD=∠FAE
在△ADB和△AEC中
∠BAD=∠FAE
∠ABD=∠ACE
BD=CE
所以△ADB≌△AEC
所以AB=AC
∵∠BAC=∠DAE
∠BAC=∠BAD+∠DAC
∠DAE=∠EAC+∠DAC
∴∠BAD=∠EAC
又∵∠ABD=∠ACE
BD=CE
∴△ABD≌△ACE(AAS)
∴AB=AC
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