证明;OA,OB,OC平分三角形ABC的各内角.则:∠OAB=(1/2)∠BAC;∠OBA=(1/2)∠ABC;
∴∠BOD=∠OAB+∠OBA=(1/2)(∠BAC+∠ABC)=(1/2)(180°-∠ACB)=90°-(1/2)∠ACB;
又OE垂直BC,则:∠COE=90° -∠OCE=90° -(1/2)∠ACB.
所以,∠BOD=∠COE.