Is there a possible geometric method to find length of this equilateral triangle?

Since $$\angle EDF = {1\over 2}\angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.

If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$

so $CE = \sqrt{13}$.

Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=\sin(\angle EDP )\cdot|ED|=1$.
We also find that $\triangle EPF$ is congruent to $\triangle CHE\ $ implying that $$ |EP|=|CH|=1. $$ By Pythagorean theorem, it follows $$ |EC|^2 =|EH|^2 +|CH|^2 =13, $$i.e. $$ |EF|=|EC| =\sqrt{13}. $$

I like the following way.

Let $\vec{AB}=\vec{a}$, $\vec{AD}=\vec{b}$, $\vec{BF}=p\vec{BD}$ and $\vec{BC}=k\vec{AD}.$

Thus, $$\vec{FE}=-p(-\vec{a}+\vec{b})-\vec{a}+\frac{1}{2}\vec{b}=(p-1)\vec{a}+\left(\frac{1}{2}-p\right)\vec{b}$$ and $$\vec{FC}=-p(-\vec{a}+\vec{b})+k\vec{b}=p\vec{a}+(k-p)\vec{b}.$$

Now, we obtain the following system: $$|\vec{FE}|=|\vec{FC}|$$ and $$\frac{\vec{FE}\cdot \vec{FC}}{|\vec{FE}||\vec{FC}|}=\frac{1}{2}$$ with variables $p$ and $k$.

We can solve this system and the rest is smooth.