证明:作AF⊥BC于F,
∵AD=AE
∴∠FAD=∠FAE(三线合一)
又∵∠BAD=∠CAE
∴∠FAD+∠BAD=∠FAE+∠CAE
即∠BAF=∠CAF
又∵AF=AF,∠BFA=∠CFA=90°
∴△BAF≌△CAF
∴AB=AC