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如图甲,直角梯形\(ABCD\)中,\(AB/\!/CD\),\(∠DAB= \dfrac {π}{2}\),点\(M\)、\(N\)分别在\(AB\),\(CD\)上,且\(MN⊥AB\),\(MC⊥CB\),\(BC=2\),\(MB=4\),现将梯形\(ABCD\)沿\(MN\)折起,使平面\(AMND\)与平面\(MNCB\)垂直\((\)如图乙\()\).
\((1)\)求证:\(AB/\!/\)平面\(DNC\);
\((2)\)当\(DN\)的长为何值时,二面角\(D-BC-N\)的大小为\(30^{\circ}\)?