如图,\(AB\)是\(⊙O\)的直径,弦\(CD⊥AB\)于点\(G\),点\(F\)是\(CD\)上一点,且满足\( \dfrac {CF}{FD}= \dfrac {1}{3}\),连接\(AF\)并延长交\(⊙O\)于点\(E\),连接\(AD\),\(DE\),若\(CF=2\),\(AF=3.\)给出下列结论:
\(①\triangle ADF\)∽\(\triangle AED\); \(②FG=2\);\(③\tan ∠E= \dfrac { \sqrt {5}}{2}\); \(④S_{\triangle DEF}=4 \sqrt {5}\)
其中正确的是\((\) \()\)