Question

已知三角形ABC,过三个顶点做角平分线交O点,AO交BC于D点,过O作BC的垂线交E点,求证＜BOD=＜COE

Answer 1

证明：∵AD平分∠BAC∴∠BAD＝∠BAC/2∵BO平分∠ABC∴∠ABO＝∠ABC/2∴∠BOD＝∠BAD+∠ABO＝（∠BAC+∠ABC）/2＝（180-∠ACB）/2＝90-∠ACB/2∵CO平分∠ACB∴∠BCO＝∠ACB/2∵OE⊥BC∴∠COE+∠BCO＝90∴∠COE＝90-∠...