Question

如图，点O、A、B在同一条直线上，OC平分∠AOD，OE平分∠FOB，∠COF=∠DOE=90°，求∠AOD

Answer 1

解：设∠COD＝X
∵∠COF＝∠DOE＝90, ∠COF＝∠COD+∠DOF，∠DOE＝∠EOF+∠DOF
∴∠EOF＝∠COD＝X
∵OC平分∠AOD
∴∠AOD＝2∠COD＝2X
∵OE平分∠FOB
∴∠BOF＝2∠EOF＝2X
∵∠AOD+∠DOF+∠BOF＝180
∴4X+∠DOF＝180
∴8X+2∠DOF＝360
∴6X+2（X+∠DOF）＝360
∴6X+2（∠COD+∠DOF）＝360
∴6X+180＝360
X＝30
∴∠AOD＝2X＝60°

Answer 2

∵∠COF=∠DOE=90°
∴∠COD=∠EOF
∵OC平分角AOD，OE平分角FOB
∴∠COD=∠AOC,∠EOF=∠BOE
∴∠COD=∠AOC=∠EOF=∠BOE
∴∵∠COF=90°
∴∠AOC+∠EOF+∠BOE=90°
∴∠AOC=30°
∴AOD=2×30°=60°

Answer 3

∠DOE=∠EOF+∠FOD=90°
∠COF=∠COD+∠FOD=90°,所以∠COD=∠EOF
所以∠EOF+∠BOE+∠AOC=90°又因为∠EOF=∠BOE=∠AOC
所以∠BOE=∠AOC=∠EOF=30°
所以∠AOD=60°

Answer 4

∠DOE=∠EOF+∠FOD=90°
∠COF=∠COD+∠FOD=90°,所以∠COD=∠EOF
所以∠EOF+∠BOE+∠AOC=90°又因为∠EOF=∠BOE=∠AOC
所以∠BOE=∠AOC=∠EOF=30°
所以∠AOD=60°