如图,分别以直角\(\triangle ABC\)的斜边\(AB\),直角边\(AC\)为边向\(\triangle ABC\)外作等边\(\triangle ABD\)和等边\(\triangle ACE\),\(F\)为\(AB\)的中点,\(DE\)与\(AB\)交于点\(G\),\(EF\)与\(AC\)交于点\(H\),\(∠ACB=90^{\circ}\),\(∠BAC=30^{\circ}.\)给出如下结论:
\(①EF⊥AC\);\(②\)四边形\(ADFE\)为菱形;\(③AD=4AG\);\(④FH= \dfrac {1}{4}BD\);
其中正确结论的是\((\) \()\)