如图,在三角形ABC中,
2个回答

连接AO并延长交BC于点E

∵∠BOE=∠BAO+∠ABO

∠COE=∠CAO+∠ACO

又∵BO,CO是角平分线

∴∠ABO=∠ABC/2

∠ACO=∠ACO/2

∴∠BOC=∠BOE+∠COE

=∠BAO+∠ABC/2 +∠CAO+∠ACB/2

=∠A+(∠ABC+∠ACB)/2

∵∠A+∠ABC+∠ACB=180

∴∠ABC+∠ACB=180-∠A

∴∠BOC=∠A+(180-∠A)/2

=90+∠A/2