如图,线段\(AB\)为\(⊙O\)的直径,点\(C\),\(E\)在\(⊙O\)上,\( \overparen {BC}= \overparen {CE}\),\(CD⊥AB\),垂足为点\(D\),连接\(BE\),弦\(BE\)与线段\(CD\)相交于点\(F\).
\((1)\)求证:\(CF=BF\);
\((2)\)若\(\cos ∠ABE= \dfrac {4}{5}\),在\(AB\)的延长线上取一点\(M\),使\(BM=4\),\(⊙O\)的半径为\(6.\)求证:直线\(CM\)是\(⊙O\)的切线.