Question

如图，直线AB，CD相交于点O，OE丄AB于O，且∠DOE=4∠COE,求∠AOD的度数。

Answer 1

解：延长EO到F
∵∠COE=∠DOF=90°-∠BOD
∴∠DOE=∠BOD+∠BOE=90°-∠BOD+90°=180°-∠BOD
又∠DOE=4∠COE
从而 180°-∠BOD=4(90°-∠BOD)=360°-4∠BOD
3∠BOD=180°
∴∠BOD=60°
从而∠COE=90°-∠BOD=90°-60°=30°
∴∠AOD=∠AOF+∠DOF=∠AOF+∠COE
=90°+30°
=120°

Answer 2

解：延长EO到F
∵∠COE=∠DOF=90°-∠BOD
∴∠DOE=∠BOD+∠BOE=90°-∠BOD+90°=180°-∠BOD
又∠DOE=4∠COE
从而 180°-∠BOD=4(90°-∠BOD)=360°-4∠BOD
3∠BOD=180°
∴∠BOD=60°
从而∠COE=90°-∠BOD=90°-60°=30°
∴∠AOD=∠AOF+∠DOF=∠AOF+∠COE
=90°+30°
=120°