如图，\(\triangle OAB\)是等腰三角形，\(∠AOB=120^{\circ}.\)以\(O\)为圆心，\(\dfrac{1}{2}\)\(OA\)为半径作圆．

\((I)\)证明：直线\(AB\)与\(⊙O\)相切

\((II)\)点\(C\)，\(D\)在\(⊙O\)上，且\(A\)，\(B\)，\(C\)，\(D\)四点共圆，证明：\(AB/\!/CD\)．