Proving $\triangle DFG$ is equilateral

Question

In this figure $\Delta ADC,\Delta BED$ are equilateral. $F$ is the midpoint of $AE$,$G$ is the midpoint of $CB$.Prove that $\Delta DFG$ is also equilateral.

My Try

Easily we can show that $$\Delta ADE \cong \Delta CDB \tag{SAS}$$ hence $$AE=BF$$ $$AF=FE=CG=BG$$ The rest is getting complicated...

Answer 1

Nope, the rest is elementary.

We have $\triangle ADF \cong \triangle CDG$ (SAS), since $AF = CG, AD = DC$ and $\angle EAD = \angle BCD$ from the pair of congruent triangles you showed. This gives $DF = DG$.

Finally $\angle FDG = \angle CDG - \angle CDF = \angle ADF -\angle CDF = 60^\circ$.

(Essentially, $\triangle CDG$ is $\triangle ADF$ rotated by $60^\circ$.)

Two equal sides with an included angle of $60^\circ$ produces an equilateral triangle.

Answer 2

Rotate everything with respect to $D$ for $60^\circ$ in counterclockwise direction.

1. Since $\triangle BDE$ is equilateral, $B$ get rotated into $E$.
2. Since $\triangle DCA$ is equilateral, $C$ get rotated into $A$.

Since rotation preserve midpoint-ness, $G$ (midpoint of $BC$) get rotated into $F$ (midpoint of $EA$). This implies $DG = DF$ and $\angle GDF = 60^\circ$. So $\triangle DFG$ is equilateral.