题目如图,⊙O\odot O⊙O是△ABC\triangle ABC△ABC的外接圆,ABABAB是⊙O\odot O⊙O的直径,点DDD是⊙O\odot O⊙O上的点,连接ADADAD,ODODOD,若ABABAB平分∠CAD\angle CAD∠CAD,∠B=72∘\angle B=72^{\circ}∠B=72∘,则∠BOD\angle BOD∠BOD的度数为( )A.18∘18^{\circ}18∘B.36∘36^{\circ}36∘C.54∘54^{\circ}54∘D.72∘72^{\circ}72∘知识点:圆周角定理I章节:图形的性质 / 圆