Question

如图所示,已知角MON=90度,点A,B分别在射线OM,ON上移动,角OAB的内角平分线与角OBA的外角平分线所在直线相

Answer 1

∠ACB＝45°,不会随着A、B点移动而改变证明：∵∠MON＝90∴∠OBE＝∠MON+∠OAB＝90+∠OAB∵BC平分∠OBE∴∠EBC＝∠OBE/2＝45+∠OAB/2∵AC平分∠OAB∴∠BAC＝∠OAB/2∴∠EBC＝∠ACB+∠BAC＝∠ACB+∠OAB/2∴∠ACB+∠OA...