Question

如图,已知▲ABC的三条角平分线交于点O作OD⊥BC于点D.求证∠COE＝∠BOD

Answer 1

∠BOE是ΔABO的外角,∴∠BOE=∠OAB+∠OBA,

∵OA、OB平分∠BAC,∠ABC,

∴∠BOE=1/2(∠BAC+∠ABC)=1/2(180°-∠ACB)=90°-1/2∠ACB,

∵OC平分∠ACB,∴∠OCD=1/2∠ACB,

∵OD⊥BC,∴∠COD=90°-∠OCD=90°-1/2∠ACB,

∴∠BOE=∠COD,

∴∠BOE+∠DOE=∠COD+∠DOE,

即∠BOD=∠COE.