Question

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

要用X算，要用因为所以写清楚，每个步骤后边括号里的依据写清楚，好的追加

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°

Answer 2

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°