∵角ABC 角ACB的平分线相交于O点
∴∠OBC=∠ABC/2,∠OCB=∠ACB/2,
∴∠OBC+OCB=∠ABC/2+∠ACB/2=(1/2)(∠ABC+∠ACB)=(1/2)(180-∠A)=90-∠A/2
即角BOC=90°+½角A