Question

如图,在三角形ABC中,BO为∠ABC的平分线,CO为三角形外角∠ACD的角平分线BO,CO交于点O,则∠BOC与∠A之间有何

Answer 1

∠A=2∠BOC
证明:
∵∠ACD=∠A+∠ABC，CO平分∠ACD
∴∠OCD＝∠ACD/2＝（∠A+∠ABC）/2
∵BO平分∠ABC
∴∠OBC＝∠ABC/2
∴∠OCD＝∠BOC+∠OBC＝∠BOC+∠ABC/2
∴(∠A+∠ABC)/2=∠BOC+∠ABC/2
∴∠A=2∠BOC

Answer 2

No picture you say a JB.