Question

如图,直线AB,CD相交于点O,OE平分∠BOD,∠AOD:∠BOE=8:1，求∠AOC的度数

Answer 1

∵,OE平分∠BOD
∴∠BOE=1/2∠BOD
∵,∠AOD:∠BOE=8:1
∴∠AOD∶1/2∠BOD=8∶1
∠AOD∶BOD=4∶1
∠AOD=4∠BOD
∵∠AOD+∠BOD=180°
∴5∠BOD+∠BOD=180°
∠BOD=36°
∴∠AOC==∠BOD=36°

追问

如图，直线AB、CD相交于点O，OE平分∠AOD，∠COF=90°，∠BOF=40°，求∠AOC、∠AOE的度数

追答

∵∠AOC+∠COF+∠BOF=180°
∴∠AOC=180°-∠COF-∠BOF=180°-90°-40°=50°
∵∠BOD=∠AOC=50°
∴∠AOD=180°-∠BOD=180°-50°=130°
∵OE平分∠AOD
∴∠AOE=1/2∠AOD=1/2×130°=65°