Find segment ratio when three points form three lines dividing area of equilateral triangle evenly $\triangle{ABC}$ is equilateral. Points $E,F,G$ are inside it and form four triangles so that $S_1=S_2=S_3=S_4$. Find $\dfrac{AF}{FG}$.

I saw this math problem on social media but did not see a strict solution. I had some results after applying barycentric coordinates to the points but did not get the final result: if mark the area of $\triangle{ACE}, \triangle{BFA}, \triangle{CGB}, \triangle{ABC}$ as $S_4, S_5, S_6, S$, then
$(S-(6+2\sqrt{5})S_4)(S-(6-2\sqrt{5})S_5)+(S-(6+2\sqrt{5})S_5)(S-(6-2\sqrt{5})S_6)+(S-(6+2\sqrt{5})S_6)(S-(6-2\sqrt{5})S_1)=0$
Maybe I need some strict proof to show the symmetry for these triangles.
Thanks.

Join G to C , we have:
$\frac{A_{GFC}}{A_{AFC}}=\frac{FG}{FA}=\frac{EG}{GB}$
$A_{GFC}=A_{GFE}+A_{GEC}$
Putting this in first relation we get:
$\frac{FG}{FA}=\frac{EG}{GB}=1+\frac{EG}{EB=EG+GB}=1+\frac{FG}{GA=FA+FG}$
Rearranging we get:
$\frac{FG}{FA}-1=\frac{FG}{FA+FG}$
Or: $FG^2-FA^2=FA\times FG$
Let $FG=x$ and $FA=y$, we have:
$x^2-y^2-xy=0$
Solving for x we find:
$x=\frac{y\pm\sqrt{y^2+4y^2}}{2}$
$\frac xy=\frac {FG}{FA}=\frac{1+\sqrt 5}2$
