(1)∵BD平分∠ABC,
∴∠ABD=∠CBD,
∵EF∥BC,
∴∠EDB=∠DBC,
∴∠ABD=∠EDB,
∴BE=ED,
同理DF=CF,
∴BE+CF=EF;
(2)BE-CF=EF,
由(1)知BE=ED,
∵EF∥BC,∴∠EDC=∠DCG=∠ACD,
∴CF=DF,
又∵ED-DF=EF,
∴BE-CF=EF.
(1)∵BD平分∠ABC,
∴∠ABD=∠CBD,
∵EF∥BC,
∴∠EDB=∠DBC,
∴∠ABD=∠EDB,
∴BE=ED,
同理DF=CF,
∴BE+CF=EF;
(2)BE-CF=EF,
由(1)知BE=ED,
∵EF∥BC,∴∠EDC=∠DCG=∠ACD,
∴CF=DF,
又∵ED-DF=EF,
∴BE-CF=EF.