Question

如图所示，直线AB、CD相交于点O，OE平分∠AOC,∠AOD-∠AOE=60°，求∠AOD的度数

Answer 1

因为OE平分∠AOC，所以∠AOD+∠AOE+∠EOC=∠AOD+2∠AOE=180°，
又因为∠AOD-∠AOE=60°，所以∠AOE=∠AOD-60°，
把∠AOE=∠AOD-60°代入∠AOD+∠AOE=180°，
得∠AOD=120°

Answer 2

∵直线AB、CD相交于点O，OE平分∠AOC
∴∠AOE=∠COE
∴∠AOE+∠COE+∠AOD=180°
∴2∠AOE+∠AOD=180°
又∵∠AOD-∠AOE=60°
∴∠AOD=100°

Answer 3

∠AOD=180-∠AOC=180-2∠AOE
又,∠AOD-∠AOE=60
所以,∠AOE=40 ,
∠AOD=100

Answer 4

直线AB、CD相交于点O，OE平分∠AOC
∠AOE=∠COE
∠AOE+∠COE+∠AOD=180°
2∠AOE+∠AOD=180°
又∵∠AOD-∠AOE=60°
∠AOD=100°