试题分析:(1)根据角平分线的性质可得∠DOC ∠AOC,∠COE ∠BOC,则∠DOE=∠DOC+∠COE ∠AOC ∠BOC ∠AOB,即可求得结果; (2)根据角平分线的性质可得∠DOC ∠AOC,∠COE ∠BOC,则∠DOE=∠DOC-∠COE ∠AOC ∠BOC ∠AOB,即可求得结果. (1)∵OD、OE分别平分∠AOC、∠COB ∴∠DOC ∠AOC,∠COE ∠BOC ∴∠DOE=∠DOC+∠COE ∠AOC ∠BOC ∠AOB ; (2)∵OD、OE分别平分∠AOC、∠COB ∴∠DOC ∠AOC,∠COE ∠BOC ∴∠DOE=∠DOC-∠COE ∠AOC ∠BOC ∠AOB![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021112638-31829.png) 点评:解题的关键是熟练掌握角的平分线把角分成相等的两个小角,且都等于大角的一半,注意本题要有整体意识. |