题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,∠ABC=90∘\angle ABC=90^{\circ}∠ABC=90∘,∠A=32∘\angle A=32^{\circ}∠A=32∘,点BBB、CCC在⊙O\odot O⊙O上,边ABABAB、ACACAC分别交⊙O\odot O⊙O于DDD、EEE两点,点BBB是CD^\widehat {CD}CD的中点,则∠ABE=______.\angle ABE=\_\_\_\_\_\_.∠ABE=______.知识点:切线的判定章节:图形的性质 / 圆