Question

如图所示,已知OD是∠AOB内的一条射线,OE平分∠AOD,OF平分∠BOD,若∠EOF=25°,求∠AOB的度数

Answer 1

∵OE平分∠AOD,OF平分∠BOD∴∠AOE=∠DOE，∠BOF=∠DOF
∵∠EOF=∠DOE+∠DOF=25°
∴∠AOD+∠BOD=2∠EOF=∠AOB=50°
望采纳，谢谢

Answer 2

证明：∵OE平分∠AOD,OF平分∠BOD
∴∠AOE=∠DOE=1/2∠AOD，∠BOF=∠DOF=1/2∠BOD
∵∠EOF=∠DOE+∠DOF=1/2∠AOD+1/2∠DOB=1/2∠AOD+∠DOB）=1/2∠AOB=25°
∴∠AOB=50°

Answer 3

50°

追问

能写下过程吗，谢谢

追答

∠EOF=∠EOD+∠DOF=1/2∠AOD+1/2DOB=1/2∠AOB=25
所以∠AOB=50