Question

已知，如图，△ABC的三个内角平分线交于O点，过O作OE⊥BC于E。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-∠BCO＝90-∠ACB/2

∴∠BOD＝∠COE