如下图,等腰三角形ABC中,AB=AC,D为CB延长线上一点,E为BC延长线上一点,且满足AB2=DB·CE.

1证明：∵AB2=DB·CE
∴AB:DB=CE:AB
∵AB=AC
∴AC:DB=CE:AB
又∵∠DBA=180°-∠ABC=180°-∠ACB=∠ACE
∴△ADB∽△EA
2.∵△ADB∽△EAC
∴∠ADB=∠CAE
∵AB=AC
∴∠ABC=∠ACB
∴∠ABC=（180°-∠BAC）/2=70°
∵∠ADB+∠DAB=∠ABC=70°
∴∠DAE=∠DAB+∠BAC+∠ADB=40°+70°=110°