\(\left\{{}\begin{matrix}\widehat{DAB}=\widehat{ABC}\left(hthang.cân\right)\\AD=BC\left(hthang.cân\right)\\AB.chung\end{matrix}\right.\Rightarrow\Delta ADB=\Delta BCA\left(c.g.c\right)\\ \Rightarrow\widehat{ADB}=\widehat{ACB}\Rightarrow\widehat{ADC}-\widehat{ADB}=\widehat{BCD}-\widehat{ACB}\\ \Rightarrow\widehat{OCD}=\widehat{ODC}\\ \Rightarrow\Delta COD.cân.tại.O\\ Mà.\widehat{COD}=\widehat{AOB}=60^0\Rightarrow\Delta COD.đều\)

Mà DF là trung tuyến nên cũng là đường cao

Do đó \(DF\perp AC\)

\(\Delta DFA\) có FM là trung tuyến ứng với cạnh huyền nên \(FG=\dfrac{1}{2}AD\left(1\right)\)

Cmtt \(\Rightarrow\Delta OAB.đều\Rightarrow AE\perp BD\Rightarrow EG=\dfrac{1}{2}AD\left(2\right)\)

\(\left\{{}\begin{matrix}OE=EB\\OF=FC\end{matrix}\right.\Rightarrow EF\) là đtb tam giác OBC \(\Rightarrow EF=\dfrac{1}{2}BC=\dfrac{1}{2}AD\left(hthang.cân\right)\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow FG=EG=EF\Rightarrow\Delta GEF.đều\)