题目如图,在△ABC\triangle ABC△ABC中,AB=ACAB=ACAB=AC,点DDD、EEE分别在边BCBCBC、ACACAC上(均不与点AAA、BBB、CCC重合),且∠1=∠C=40∘\angle 1=\angle C=40^{\circ}∠1=∠C=40∘,若BD=CEBD=CEBD=CE,则∠DAB=\angle DAB=∠DAB=______度.知识点:角的运算章节:未标注