Question

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

Answer 1

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°