∵∠ABC和∠ACB的平分线BD，CE相交于点O，∠ABC=40°，∠ACB=80°，∴∠DBC= 1 2 ∠ABC=20°，∠ECB= 1 2 ∠ACB=40°，∴∠BOC=180°-∠DBC-∠ECB=180°-20°-40°=120°．答：∠BOC=120°．
