(1)∵BD=BA
∴∠BAD=∠ABD=(180-∠B)/2=67.5°
又∠ADB=∠DAE+∠E
∴∠DAE=67.5-22.5=45°
(2)∵CE=CA
∴∠E=∠CAE=x
则∠ACB=2x
∴∠B=90-2x
∠ADB=[180-(90-2x)]/2=45+x
又∠ADB=∠DAE+∠E
∴∠DAE=45+x-x=45°
∴不变
(3)∠BAC=2∠DAE
(1)∵BD=BA
∴∠BAD=∠ABD=(180-∠B)/2=67.5°
又∠ADB=∠DAE+∠E
∴∠DAE=67.5-22.5=45°
(2)∵CE=CA
∴∠E=∠CAE=x
则∠ACB=2x
∴∠B=90-2x
∠ADB=[180-(90-2x)]/2=45+x
又∠ADB=∠DAE+∠E
∴∠DAE=45+x-x=45°
∴不变
(3)∠BAC=2∠DAE