Question

已知直线AB、CD,相交于点O,OE平分角BOD,OF平分角COE,角AOD比角DOE=4比1,求角AOF的度数

Answer 1

∵OE平分∠BOD ∴∠BOE=∠EOD 又∵OF平分∠COE ∴∠COF=∠FOE ∵∠AOD:∠DOE=4:1 ∴设∠EOD=X 则∠AOD=4X 又∵∠BOE=∠EOD ∴∠BOE=∠EOD=X 4X+2X=180 6X=180 X=30 ∴∠BOE=∠EOD=30° ∵∠COF=∠FOE ∴设∠COF=∠FOE=y ∴2y+30=180 2y=150 y=75 ∴∠COF=∠FOE=75° ∴∠FOB-∠BOE=75-30=45° ∴∠AOF=180-45=135°