方法一：证明：由OE⊥BC可知：∠BCO与∠COE互余:∠COE=90°-∠BCO又∠ABC+∠ACB+∠BAC=180°∵O是三角形ABC内心,∴∠BCO+∠BAO+∠ABO=90°∠BAO+∠ABO=∠BOD即∠BCO与∠BOD互余:∠BOD=90°-∠BCO∴∠BOD＝∠COE=90°-∠BCO∴∠BOD＝∠COE方法二：证明：∠BOD=∠ABO+∠BA0=（∠BAC+∠ABC）/2=（180°-∠ACB）/2∠BOD=90°-∠HCB∵OE⊥BC,∴∠COE=90°-∠HCB∴∠BOD=∠COE