如图\(1\),在梯形\(ABCD\)中,\(AB/\!/CD\),\(\angle ABC={{90}^{\circ }}\),\(AB=2CD=2BC=4\),\(O\)是边\(AB\)的中点\(.\) 将三角形\(AOD\)绕边\(OD\)所在直线旋转到\({{A}_{1}}OD\)位置,使得\(\angle {{A}_{1}}OB={{120}^{\circ }}\),如图\(2.\) 设\(m\)为平面\({{A}_{1}}DC\)与平面\({{A}_{1}}OB\)的交线.
\((\)Ⅰ\()\)判断直线\(DC\)与直线\(m\)的位置关系并证明;
\((\)Ⅱ\()\)若直线\(m\)上的点\(G\)满足\(OG\bot {{A}_{1}}D\),求出\({{A}_{1}}G\)的长;
\((\)Ⅲ\()\)求直线\({{A}_{1}}O\)与平面\({{A}_{1}}BD\)所成角的正弦值.\(\dfrac{\sqrt{5}}{5}\)