如图,在三棱柱\(ABC−{{A}_{1}}{{B}_{1}}{{C}_{1}}\)中,侧面\(AB{{B}_{1}}{{A}_{1}}\)是矩形,\(∠BAC=90^{\circ}\),\(A{{A}_{1}}⊥BC\),\(A{{A}_{1}}=AC=2AB=4\),且\(B{{C}_{1}}⊥{{A}_{1}}C\).
![](https://www.ebk.net.cn/tikuimages/2/2018/700/shoutiniao91/220c10bf2bb6ef401f475ad802fac145.png)
\((1)\)求证:平面\(AB{{C}_{1}}⊥\)平面\({{A}_{1}}AC{{C}_{1}}\);
\((2)\)设\(D\)是\({{A}_{1}}{{C}_{1}}\)的中点,判断并证明在线段\(B{{B}_{1}}\)上是否存在点\(E\),使得\(DE/\!/\)平面\(AB{{C}_{1}}.\)若存在,求二面角\(E−A{{C}_{1}}−B\)的余弦值.