解:∵两直线AB,CD相交于点O,∠COF=2∠AOC,
∠AOC=∠BOD
又∵OE平分∠BOD,
∴ ∠AOC=∠BOD=2∠BOE.
∵OF⊥OE
∠BOE+∠AOF=180°-∠EOF=90°=∠DOE+∠DOF
∴∠AOF=∠DOF
∵∠COF=2∠AOC
即∠BOF=2∠BOD=∠BOD+∠DOF,
∠BOD=∠DOF=∠AOF
∠BOD+∠DOF+∠AOF=180°
∠BOF=∠BOD+∠DOF=120°
追问写的真好!谢谢啦
追答不客气!