Question

如图,已知直线AB,CD交于O点,OE平分∠BOD,OF平分∠COE,∠AOD:∠BOE=4:1,求∠AOF.

Answer 1

∵∠AOD：∠BOE=4:1

∴设∠AOD=4x,则∠BOE=x

∵OE平分∠BOD,

∴∠DOE=∠BOE=x,∠BOD=2x

∠AOD+∠BOD=180°

4x+2x=180°

x=30°

∴∠BOD=2x=60°,∴∠AOC=∠BOD=60°

∠BOC=4x=120°,

∠COE=120°+30°=150°

∵OF平分∠COE,

∴∠COF=75°

∴∠AOF=∠AOC+∠COF=60+75=135°