Question

如图，直线AB，CD相交于点O，OE平分∠BOD，OF平分∠COE，且∠1：∠2=1：4，求∠AOF的度数

Answer 1

因为OE平分∠BOD，∠1=∠DOE，又因为，∠2：∠1=4:1，∠2+∠1+∠DOE=180
所以∠1=30，∠2=120=∠BOC
又OF平分∠COE，∠COE=150
∠EOF=75，∠BOF=45
∠AOF=180-∠BOF=135度

Answer 2

OE平分∠BOD
∠BOD=2∠BOE
∠AOD:∠BOE=4:1
∠AOD:∠BOD=4:2=2:1
∠AOD+∠BOD=180°
∠AOD=120° ∠BOD=60°
∠BOE=∠DOE=30°
∠AOC=∠BOD=60° ∠AOD=∠BOC=120°
OF平分∠COE
∠COF=∠BOF=120°/2=60°
∠AOF=∠AOC+∠COF=60°+60°=120°

Answer 3

OE平分∠BOD ∠
∠BOD=2∠BOE
∠AOD:∠BOE=4:1
∠AOD:∠BOD=4:2=2:1
∠AOD+∠BOD=180°
∠AOD=120° ∠BOD=60°
∠BOE=∠DOE=30°
∠AOC=∠BOD=60° ∠AOD=∠BOC=120°
OF平分∠COE
∠COF=∠BOF=120°/2=60°
∠AOF=∠AOC+∠COF=60°+60°=120°

Answer 4

图在哪里