如图,在\(\triangle ABC\)中,\(∠C=90^{\circ}\),\(AC=BC=2\),点\(D\),\(E\)分别在边\(BC\),\(AB\)上,连接\(AD\),\(ED\),且\(∠BDE=∠ADC.\)过\(E\)作\(EF⊥AD\)交边\(AC\)于点\(F\),连接\(DF\).
\((1)\)求证:\(∠AEF=∠BED\);
\((2)\)过\(A\)作\(AG/\!/ED\)交\(BC\)的延长线于点\(G\),设\(CD=x\),\(CF=y\),求\(y\)与\(x\)之间的函数关系式;
\((3)\)当\(\triangle DEF\)是以\(DE\)为腰的等腰三角形时,求\(CD\)的长.