1.
在三棱柱\(ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}\)中,侧面\(AB{{B}_{1}}{{A}_{1}}\)为矩形,\(AB=2\),\(A{{A}_{1}}=2\sqrt{2}\),\(D\)是\(A{{A}_{1}}\)的中点,\(BD\)与\(A{{B}_{1}}\)交于点\(O\),且\(CO\bot \)平面\(AB{{B}_{1}}{{A}_{1}}\).
\((1)\)证明:\(BC\bot A{{B}_{1}}\);
\((2)\)若\(OC=OA\),求直线\(CD\)与平面\(ABC\)所成角的正弦值.