Question

如图,直线AB,CD相交于点O,OE平分角BOD,OF平分角AOD,角BOC：角DOE=6:1,求（1）角DOE的度数（

Answer 1

∵OE平分∠BOD

∴∠DOE=∠BOE

∵∠BOC+∠DOE+∠BOE=180°

∠BOC：∠DOE=6：1

即8∠DOE=180° ∠DOE=22.5°

∵∠AOD+∠DOE+∠BOE=180°

∠DOE=∠BOE=22.5°∠BOC=135°

∴∠AOD=135°

∵OF平分 ∠AOD

∴∠AOF=∠DOF=67.5°

∴∠DOE=∠AOF+∠AOC

∠AOC=180°-∠BOC=45°

∴∠COF=∠AOF+∠AOC=112.5°