已知棱台\(ABC-A_{1}B_{1}C_{1}\)，平面\(AA_{1}C_{1}C⊥\)平面\(A_{1}B_{1}C_{1}\)，\(∠B_{1}A_{1}C_{1}=60°\)，\(∠A_{1}B_{1}C_{1}=90°\)，\(AA_{1}=AC=CC_{1}=\dfrac{A_{1}C_{1}}{2}\)，\(D\)，\(E\)分别是\(BC\)和\(A_{1}C_{1}\)的中点
\((I)\)证明：\(DE⊥B_{1}C_{1}\)；
\((Ⅱ)\)求\(DE\)与平面\(BCC_{1}B_{1}\)所成角的余弦值