Question

如图，直线ab，cd相交于点o，oe平分∠bod，of平分∠cow，∠aod:∠boe=7:1，求

如图，直线ab，cd相交于点o，oe平分∠bod，of平分∠cow，∠aod:∠boe=7:1，求∠aof的度数

Answer 1

解：∵∠AOD：∠BOE＝7:1
∴∠AOD=7∠BOE
又OE平分∠BOD
∴∠BOD=2∠BOE
又∠AOD+∠BOD=180°
∴7∠BOE+2∠BOE=180°
解之得：∠BOE=20°
∴∠BOD=2*20°
=40°
∴∠BOC=180°-∠BOD
=180°-40°
=140°
又OF平分∠COE
∴∠COF=(∠BOC+20°)/2
=(140°+20°)/2
=80°
∴∠BOF=80°-20°
=60°
又∠AOF+∠BOF=180°
∴∠AOF=180°-∠BOF
=180°-60°
=120°