如图\(1\),在\(\triangle ABC\)中,\(D\),\(E\)分别为\(AB\),\(AC\)的中点,\(O\)为\(DE\)的中点,\(AB=AC=2 \sqrt {5}\),\(BC=4.\)将\(\triangle ADE\)沿\(DE\)折起到\(\triangle A_{1}DE\)的位置,使得平面\(A_{1}DE⊥\)平面\(BCED\),如图\(2\).
\((\)Ⅰ\()\)求证:\(A_{1}O⊥BD\);
\((\)Ⅱ\()\)求直线\(A_{1}C\)和平面\(A_{1}BD\)所成角的正弦值;
\((\)Ⅲ\()\)线段\(A_{1}C\)上是否存在点\(F\),使得直线\(DF\)和\(BC\)所成角的余弦值为\( \dfrac { \sqrt {5}}{3}\)?若存在,求出\( \dfrac {A_{1}F}{A_{1}C}\)的值;若不存在,说明理由.