如图,已知:\(AB\)是\(⊙O\)的直径,点\(C\)在\(⊙O\)上,\(CD\)是\(⊙O\)的切线,\(AD⊥CD\)于点\(D\),\(E\)是\(AB\)延长线上一点,\(CE\)交\(⊙O\)于点\(F\),连接\(OC\)、\(AC\).
\((1)\)求证:\(AC\)平分\(∠DAO\).
\((2)\)若\(∠DAO=105^{\circ}\),\(∠E=30^{\circ}\)
\(①\)求\(∠OCE\)的度数;
\(②\)若\(⊙O\)的半径为\(2 \sqrt {2}\),求线段\(EF\)的长.