已知椭圆\(C\):\( \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1(a > b > 0)\)的离心率为\( \dfrac { \sqrt {3}}{2}\),右顶点为\(A(2,0)\).
\((\)Ⅰ\()\)求椭圆\(C\)的方程;
\((\)Ⅱ\()\)过点\((1,0)\)的直线\(l\)交椭圆于\(B\),\(D\)两点,设直线\(AB\)斜率为\(k_{1}\),直线\(AD\)斜率为\(k_{2}\),求证:\(k_{1}k_{2}\)为定值.