证明：△ABD和△ACE中,AB=AC,AD=AE,BD=CE,%D%A 故△ABD≌△ACE(SSS)%D%A ∴∠ABD=∠ACE=∠2,∠BAD=∠CAE=∠1%D%A ∠3为△ABD外角,则∠3=∠1+∠2