题目在如图所示的"箭头"模型中,已知∠ABO′=14∠ABO∠ABO′=\frac{1}{4}∠ABO∠ABO′=41∠ABO,∠ACO′=14∠ACO∠ACO′=\frac{1}{4}∠ACO∠ACO′=41∠ACO,关于结论ⅠⅠⅠ、ⅡⅡⅡ,下列判断正确的是( )结论ⅠⅠⅠ:∠BOC=∠A+∠ABO+∠ACO\angle BOC=\angle A+\angle ABO+\angle ACO∠BOC=∠A+∠ABO+∠ACO;结论ⅡⅡⅡ:若2∠BO\’C=∠BOC2\angle BO\’C=\angle BOC2∠BO\’C=∠BOC,则∠ABO+∠ACO=∠A\angle ABO+\angle ACO=\angle A∠ABO+∠ACO=∠A.A.只有结论ⅠⅠⅠ对B.只有结论ⅡⅡⅡ对C.结论ⅠⅠⅠ、ⅡⅡⅡ都对D.结论ⅠⅠⅠ、ⅡⅡⅡ都不对知识点:圆周角定理I章节:图形的性质 / 圆