Question

如图所示,O为直线AB上一点,OC为任一射线,OD平分∠AOC,OE平分∠COB.

Answer 1

(1)OD平分∠AOC,OE平分∠COB,则有∠AOD=∠COD,∠COE=∠BOE；

又因为∠AOC+∠BOC=π,则有∠DOC+∠COE=π/2.且∠COE=∠BOE.即与∠DOC互余的角为∠COE和∠BOE.

(2)因为∠AOD=∠COD,故可以找出与∠AOD互补的角,也就是与∠DOC互补的角.

如图∠AOD+∠DOE=π,即与∠DOC互补的角为∠DOE.