Question

如图所示,角ABC,角ACB的内角平分线交于点O,角ABC的内角平分线与角ACB的外角交于点D,

Answer 1

因为△ABC内角和=180°,OB平分∠ABC,OC平分∠ACB,∠A=60°

∠BOC=2倍∠A=120°

又因CD为∠ABC外角平分线,所以∠OCD=90°

∠BOC=∠OCD+∠D,所以∠D=30°

∠ABC与∠ACB的相邻外角平分线交于点E

所以∠OBE=∠OCE=90°,在四边形OBEC中,∠E+∠OBE+∠OCE+∠BOC=360°

∠E=60°