已知:如图,在三角形ABC中,BD,CE是角ABC,角ACB的角平分线,且相交于点O求证:角BOC=90度+1/2角A

∵BD平分∠ABC
∴∠DBC=∠ABC/2
∵CE平分∠ACB
∴∠ECB=∠ACB/2
∴∠DBC+∠ECB=(∠ACB+∠ABC)/2
∵∠A+∠ABC+∠ACB=180
∴∠ABC+∠ACB=180-∠A
∴∠DBC+∠DCB=90-∠A/2
∵∠BOC+∠DBC+∠ECB=180
∴∠BOC=90+∠A/2∴∠DBC+∠DCB=90-∠A/2 此处应是∠DBC+∠ECB=90-∠A/2吧。对