6.
已知\(\triangle ABC\)是等腰直角三角形,\(AC=BC=2\),\(D\)是边\(AB\)上一动点\((A\)、\(B\)两点除外\()\),将\(\triangle CAD\)绕点\(C\)按逆时针方向旋转角\(α\)得到\(\triangle CEF\),其中点\(E\)是点\(A\)的对应点,点\(F\)是点\(D\)的对应点.
\((1)\)如图\(①\),当\(α=90^{\circ}\)时,\(G\)是边\(AB\)上一点,且\(BG=AD\),连接\(GF.\)求证:\(GF/\!/AC\);
\((2)\)如图\(②\),当\(90^{\circ}\leqslant α\leqslant 180^{\circ}\)时,\(AE\)与\(DF\)相交于点\(M\).
\(①\)当点\(M\)与点\(C\)、\(D\)不重合时,连接\(CM\),求\(∠CMD\)的度数;
\(②\)设\(D\)为边\(AB\)的中点,当\(α\)从\(90^{\circ}\)变化到\(180^{\circ}\)时,求点\(M\)运动的路径长.