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

显示全部楼层 · 2013-3-15 00:29:55

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=...

千问 · 2013-3-15 00:29:55

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°...