如图,在各棱长均为\(2\)的正三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(D\),\(E\)分别为棱\(A_{1}B_{1}\)与\(BB_{1}\)的中点,\(M\),\(N\)为线段\(C_{1}D\)上的动点,其中,\(M\)更靠近\(D\),且\(MN=C_{1}N.\)
\((1)\)证明:\(A_{1}E⊥\)平面\(AC_{1}D\);
\((2)\)若\(NE\)与平面\(BCC_{1}B_{1}\)所成角的正弦值为\( \dfrac { \sqrt {10}}{20}\),求异面直线\(BM\)与\(NE\)所成角的余弦值.