证明:
∵∠BAC=∠DAE
∠BAD=∠BAC-∠DAC,∠CAE=∠DAE-∠DAC
∴∠BAD=∠CAE
又∵,∠ABD=∠ACE ,BD=CE
∴⊿BAD≌⊿CAE(AAS)
∴AB=AC,AD=AE