如图,直线AB、CD相交于点O,OE平分∠BOD,OF平分∠COE,∠AOD：∠BOE=7:1,求∠AOF的度数.

∵OE平分∠BOD,∴∠DOE=∠BOE。∵∠AOD:∠BOE=7:1,∴∠AOD=7∠BOE。又∵∠AOD+∠DOE+∠BOE=180°,∴7∠BOE+∠BOE+∠BOE=180°,∴∠DOE=∠BOE=20°。∴∠COE=180°-20°=160°。∵OF平分∠COE,∴∠EOF=1/2∠COE=80°。∴∠BOF=80°-20°=60°∴∠AOF=180°-∠BOF=120°