题目如图,ABABAB是⊙O\odot O⊙O的直径,CCC,DDD为⊙O\odot O⊙O上两点,且BDBDBD平分∠CBA\angle CBA∠CBA,连接CDCDCD,ACACAC,若∠CDB=32∘\angle CDB=32^{\circ}∠CDB=32∘,则∠ACD\angle ACD∠ACD的度数为______∘.^{\circ}.∘.知识点：切线的性质章节：图形的性质 / 圆
如图,是的直径,,为上两点,且平分,连接,,若,则的度数为______如图,是的直径,,为上两点,且平分,连接,,若,则的度数为______如图,是的直径,,为上两点,且平分,连接,,若,则的度数为______