如图所示,在正方形ABCD的边CB的延长线上取点F,连接AF,在AF上取点G,使得AG=AD,连接DG,过点A作AE⊥AF,交DG于点E.
(1)若正方形ABCD的边长为4,且tan∠FAB=
(2)求证:AE+BF=AF.