Question

如图,直线AB、CD相交于点O,OE是∠AOD的平分线,若∠AOC=60°,OF⊥OE。求∠BOE的度数

Answer 1

由已知 ∠AOC=60° => ∠AOC的对顶角∠BOD=60°
∠AOE =180° -∠BOD = 120°
由已知OE是∠AOD的平分线 => ∠DOE = 120°/2 =60°

∠BOE=∠BOD+∠DOE =60° +60° = 120°

Answer 2

∠AOC=26°
∠AOD=180°-26°=154°OE是∠AOD的平分线∠AOE=1/2∠AOD=77°

Answer 3

你的图呢？

Answer 4

图呢。