Question

在三角形ABC中,三个内角的角平分线交于点O,OE⊥BC于点E

试说明角BOD=角COE

Answer 1

证明：
∵∠BAC+∠ABC+∠ACB＝180
∴∠ACB＝180-（∠BAC+∠ABC）
∵AO平分∠BAC
∴∠BAO＝∠BAC/2
∵BO平分∠ABC
∴∠ABO＝∠ABC/2
∵∠BOD＝∠BAO+∠ABO
∴∠BOD＝∠BAC/2+∠ABC/2＝（∠BAC+∠ABC）/2
∵CO平分∠ACB
∴∠BCO＝∠ACB/2＝[180-（∠BAC+∠ABC）]/2＝90-（∠BAC+∠ABC）/2
∵OE⊥BC
∴∠COE+∠BCO＝90
∴∠COE＝90-∠BCO＝90-90+（∠BAC+∠ABC）/2＝（∠BAC+∠ABC）/2
∴∠BOD＝∠COE

Answer 2