如图,在三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,侧棱\(A_{1}A⊥\)平面\(ABC\),\(AC⊥BC\),\(AC=1\),\(BC=2\),\(S\),点\(D\)是\(AB\)的中点.
\((I)\)证明:\(AC_{1}/\!/\)平面\(CDB_{1}\);
\((\)Ⅱ\()\)在线段\(AB\)上找一点\(P\),使得直线\(AC_{1}\)与\(CP\)所成角的为\(60^{\circ}\),求\( \dfrac {| \overrightarrow{AP}|}{| \overrightarrow{AB}|}\)的值.