△ABC的内角∠ABC,外角∠ACD的角平分线OB,OC相交于点O,求∠BOC与∠A的关系

Question

△ABC的内角∠ABC,外角∠ACD的角平分线OB,OC相交于点O,求∠BOC与∠A的关系

Answer 1

∠A=180-∠ABC-∠ACB
∠BOC=180-∠OBC-∠OCB
∵∠OCB=∠ACB+∠ACO=∠ACB+1/2(180-∠ACB)=90+1/2∠ACB
又∵∠OBC=1/2∠ABC
∴∠BOC=180-（90+1/2∠ACB）-1/2∠ABC=90-1/2∠ACB-1/2∠ABC=1/2∠A