如图,直线BC、DE交于点O,AO⊥OB,OF平分∠COE,∠COF+∠BOD=51°,求∠AOD的度数

Question

如图,直线BC、DE交于点O,AO⊥OB,OF平分∠COE,∠COF+∠BOD=51°,求∠AOD的度数

Answer 1

∵∠COE和∠BOD是对角
∴∠COE=∠BOD
∵OF平分∠COE
∴∠COE=2∠COF
∴∠BOD=2∠COF
∵∠COF+∠BOD=51°
∴3∠COF=51°
∴∠COF=17°
∵AO⊥OB
∴∠AOC=90°
∴∠AOD+∠COE=90°
∴∠AOD=90°-∠COE=90°-2∠COF=90°-34°=56°