Question

如图,直线AB、CD相交于点O,OE平分∠BOD,OF平分∠COB,∠1：∠2=1：4,求∠AOF的度数.

Answer 1

其实很容易,

OE平分∠BOD,∠AOC和∠BOD角度一样,所以,

∠DOE = ∠BOE = ∠1

∠AOC = ∠BOD = 2∠1

OF平分∠COB,∠AOD和∠BOC角度一样,所以,

∠AOD = ∠EOB = ∠2

∠COF = ∠BOF = 0.5∠BOC = 0.5∠2

∠1 ：∠2 = 1 ：4

∠1 / ∠2 = 1/4

∠2 = 4∠1__i

既然COD是直线,

∠AOC + ∠AOD = 180°

2∠1 + ∠2 = 180°__ii

把 i 放入 ii,

2∠1 + 4∠1 = 180°

6∠1 = 180°

∠1 = 30°

∠AOF相等于,

∠AOF = ∠AOC + ∠EOF

= 2∠1 + 0.5∠2

= 2∠1 + 0.5(4∠1)

= 2∠1 + 2∠1

= 4∠1

= 4(30°)

= 120°

结论∠AOF = 30°