如图,在三棱柱\(ABC−{{A}_{1}}{{B}_{1}}{{C}_{1}}\)中,\(C{{C}_{1}}\bot \)平面\(ABC\),\(D\),\(E\),\(F\),\(G\)分别为\(A{{A}_{1}}\),\(AC\),\({{A}_{1}}{{C}_{1}}\),\(B{{B}_{1}}\)的中点,\(AB=BC=\sqrt{5}\),\(AC=A{{A}_{1}}=2\).
\((\)Ⅰ\()\)求证:\(AC⊥\)平面\(BEF\);
\((\)Ⅱ\()\)求二面角\(B−CD−C_{1}\)的余弦值;
\((\)Ⅲ\()\)证明:直线\(FG\)与平面\(BCD\)相交.