如图：直线AB、CD相交于O，OE⊥OF，∠BOF=2∠BOE，OC平分∠AOE，求：∠DOE的度数．

∵OE⊥OF，∠BOF=2∠BOE，∴∠BOF+∠BOE=3∠BOE=90°，解得∠BOE=30°，∴∠BOF=2×30°=60°，∴∠AOE=180°-∠BOE=150°，∵OC平分∠AOE，∴∠AOC=12∠AOE=12×150°=75°，∴∠BOD=∠AOC=75°，∠DOE=∠BOD+∠BOE...