Question

如图，已知直线AB、CD、EF相交于点O，OF平分∠BOD，∠COB＝∠AOC＋40°，求∠AOF

如图，已知直线AB、CD、EF相交于点O，OF平分∠BOD，∠COB＝∠AOC＋40°，求∠AOF的度数．

Answer 1

∵∠AOC+∠COB=180°
∴∠COB=180°-∠AOC
∵,∠COB=∠AOC+40°
∴180°-∠AOC=∠AOC+40°
∴∠AOC=70°=∠BOD
∴∠COB=110°=∠AOD
∵OF平分∠BOD
∴∠DOF=35°
∴∠AOF=∠AOD+∠DOF=110°+35°=145°

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Answer 2

由图：∠COB+∠AOC＝180°，又因：∠COB=∠AOC+45°　代入
∠AOC+45°+∠AOC＝180°，
解：∠AOC＝67.5°　　，∠COB=∠AOC+45°＝112.5°
因OF平分∠BOD，即OE平分∠AOC，∠EOC＝∠EOA＝1/2∠AOC＝33.75°
又因，∠EOC与，∠FOD对顶角所以，∠FOD＝33.75°　
∠COB与∠AOD对顶角，所以∠AOD＝∠COB＝112.5°
∠AOF＝∠AOD+∠FOD＝112.5°+33.75°　＝146.25°