(1)∵AD平分∠CAB,
∴∠EAD=∠BAD,
∵EF∥AB,
∴∠EDA=∠BAD,
∴∠EDA=∠EAD,
∴AE=ED,
∵EF=AE+BF,
∴DF=BF,
∴∠FBD=∠FDB,
∵EF∥AB,
∴∠FDB=∠DBA,
∴∠FBD=∠DBA,
∴BD平分∠CBA.
(2)①AE=BF+EF,
理由是:∵EF∥AB,
∴∠EDA=∠DAB,∠EDB=∠DBH,
∵AD平分∠CAB,BD平分∠CBH,
∴∠DAB=∠DAE,∠DBH=∠DBC,
∴∠EDA=∠DAE,∠FDB=∠CBD,
∴AE=DE,DF=BF,
∴AE=DE=EF+DF=EF+BF.
②①中的结论始终成立,
理由是:∵EF∥AB,
∴∠EDA=∠DAB,∠EDB=∠DBH,
∵AD平分∠CAB,BD平分∠CBH,
∴∠DAB=∠DAE,∠DBH=∠DBC,
∴∠EDA=∠DAE,∠FDB=∠CBD,
∴AE=DE,DF=BF,
∴AE=DE=EF+DF=EF+BF.