(1)证明：∵AE⊥面ABC,面ABC⊥面BCD,且交于BC,点M在BC上
又AM⊥BD,AM∈面ABC
∴AM⊥面BCD==>AM⊥BC
(2)设M为BC中点,AB=AC=AE=CD=BD=3,BC=3√2
∴⊿ABC≌⊿DBC,DM⊥BC,AM=MD=3√2/2
∴BD⊥DC
∴⊿AEC≌⊿AEB==BE=EC,
∴⊿CDE≌⊿BDE,
过B作BF⊥ED交ED延长线于F,连接CF,则CF⊥ED,CF=BF
则∠BFC为二面角B-DE-C的平面角
DE=√[(AE-DM)^2+AM^2]=√[(AE^2+DM^2-2AE*DM+AM^2]
=√(18-9√2]
设DF=x
BE^2-(DE+x)^2=BF^2=BD^2-x^2
BE^2-DE^2-2DEx=BD^2
X= (BE^2-DE^2-BD^2)/DE=[18-(18-9√2)-9]/
√(18-9√2]=3√(4-2√2)/2
BF^2=BD^2-x^2=9-9(4-2√2)/4=9√2/2
cos∠BFC=(BF^2+FC^2-BC^2)/(2BF*FC)=1-4/9=5/9
∴二面角B-DE-C的余弦值为5/9