Question

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

Answer 1

∵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