(1)△ABD∽△EAC,理由如下:
∵AB=AC,AB²=BD·CE,∴AB/CE=BD/AC
又∵AB=AC,∴∠ABC=∠ACB
∴∠ABD=∠ACE,∴△ABD∽△EAC
(2)∵AB=AC,∠BAC=40°,∴∠ABC=∠ACB=70°
∵△ABD∽△EAC,∴∠D=∠CAE
又∵∠DAE=∠DAB+∠BAC+∠CAE,∠D+∠DAB=∠ABC
∴∠DAE=∠ABC+∠BAC=70°+40°=110°
(1)△ABD∽△EAC,理由如下:
∵AB=AC,AB²=BD·CE,∴AB/CE=BD/AC
又∵AB=AC,∴∠ABC=∠ACB
∴∠ABD=∠ACE,∴△ABD∽△EAC
(2)∵AB=AC,∠BAC=40°,∴∠ABC=∠ACB=70°
∵△ABD∽△EAC,∴∠D=∠CAE
又∵∠DAE=∠DAB+∠BAC+∠CAE,∠D+∠DAB=∠ABC
∴∠DAE=∠ABC+∠BAC=70°+40°=110°