题目如图,AB=ADAB=ADAB=AD,点BBB关于ACACAC的对称点EEE恰好落在CDCDCD上.若∠BAD=α(0∘<a<180∘)\angle BAD=\alpha \left(0^{\circ} \lt a \lt 180^{\circ}\right)∠BAD=α(0∘<a<180∘),∠ACB=β\angle ACB=\beta∠ACB=β,则下列关系正确的是( )A.a−β=90∘a-\beta =90^{\circ}a−β=90∘B.c=3βc=3\betac=3βC.α+β=180∘\alpha +\beta =180^{\circ}α+β=180∘D.a+2β=180∘a+2\beta =180^{\circ}a+2β=180∘知识点:圆周角定理I章节:图形的性质 / 圆