如图,在四边形\(ABCD\)中,\(AB/\!/CD\),\(∠ADC=90^{\circ}\),\(DE⊥BC\)于\(E\),连\(AE\),\(FE⊥AE\)交\(CD\)于点\(F\).
\((1)\)求证:\(\triangle AED\)∽\(\triangle FEC\);
\((2)\)若\(AB=2 \sqrt {3}\),求\(DF\)的值;
\((3)\)若\(AD=CD\),\( \dfrac {S_{\triangle AEB}}{S_{\triangle DEF}}=2\),则\( \dfrac {AB}{CD}=\) ______ .