Question

已知∠ACB=60°,∠ABC=50°,BO,CO分别平分∠ABC,∠ACB,EF经过点O且平行于BC,求∠BOC的度数

答案是125°，我要过程，请快点！（有图）

Answer 1

∠EOB=∠OBC=∠ABC/2=25°
∠FOC=∠OCB=∠ACB/2=30°
∠BOC=180°-25°-30°=125°

Answer 2

∵EF∥BC
∴∠EOB=∠OBC，∠FOC=∠OCB
又∵BO，CO分别平分∠ABC，∠ACB
∴∠EBO=∠OBC=1/2∠ABC=25°，∠ACO=∠OCB=1/2∠ACB=30°
∴∠EOB=25°，∠FOC=30°
∴∠BOC＝180°-25°-30°＝125°

Answer 3

∵BO,CO分别平分∠ABC,∠ACB ∴∠OBC= ∠ABC/2=25° ∠OCB=∠ACB/2=30°
∵EF∥BC，∴ ∠EOB=∠OBC ,∠FOC=∠OCB 则∠EOB+∠FOC=25°+30°=55°
∴∠BOC=180°-55°=125°