∵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