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.
 A: Here's an argument that doesn't assume the lines are symmetrically arranged.

Consider equilateral $\triangle ABC$ of side-length $4s$, and let equilateral $\triangle A'B'C'$ (of side-length $s$) be the center instance of the sixteen-triangle subdivision of $\triangle ABC$. That is, the extended edges of $\triangle A'B'C'$ are parallel to, and quadrisect, those of $\triangle ABC$. Consequently, vertex $P$ of $\triangle PBC$, whose area is one-quarter that of $\triangle ABC$, must lie on extended edge $\overline{B'C'}$; likewise vertex $Q$ of $\triangle AQC$ and vertex $R$ of $\triangle ABR$ lie on extended edges $\overline{A'C'}$ and $\overline{A'B'}$.
Define $p:=|B'P|$, $q:=|C'Q|$, $r:=|A'R|$. Also, let the extension of $\overline{A'C'}$ meet $\overline{BC}$ at $Q'$; note that $|B'C'|=|Q'C'|=s$.

Since $\triangle QQ'C\sim\triangle QC'P$, we have
$$\frac{|QQ'|}{|Q'C|}=\frac{|QC'|}{|C'P|}\quad\to\quad\frac{q+s}{s}=\frac{q}{p-s}\quad\to\quad p = \frac{s (s + 2q)}{s+q} \tag1$$
Likewise, by situational symmetry,
$$\begin{align}
r &= \frac{s (s+2p)}{s+p}  \quad\stackrel{(1)}{=}\quad \frac{s (3s+5q)}{2s+3 q}  \tag2 \\[4pt]
q &= \frac{s (s+2r)}{s+r}  \quad\stackrel{(2)}{=}\quad \frac{s (8s+13q)}{5s+8 q}  \tag3 \\[4pt]
\end{align}$$
From $(3)$, we obtain
$$\frac{q}{s} = \frac{s+q}{q} \tag4$$
which, for positive $s$ and $q$ says that we have the Golden ratio
$$\frac{q}{s}=\phi=\frac12(1+\sqrt{5})=1.618\ldots \tag5$$
(By identical arguments, or by back-substituting into $(1)$ and $(2)$, $\frac{r}{s}=\frac{p}{s}=\phi$, as well, which gives us guarantees the geometric symmetric $p=q=r$.) But $q/s$ is, by those similar triangles, equal to the ratio $|PQ|/|CP|$, which is the reciprocal of the target ratio; thus, the target ratio itself is $$\phi^{-1}=\frac12(-1+\sqrt{5})=0.618\ldots \tag6$$
$\square$

Note: The negative solution of $(4)$ implies $\frac{q}{s}=-\phi^{-1}=1-\phi$, which corresponds to $P$ situated on the opposite side of $\overline{B'C'}$. Likewise for $Q$ and $R$, giving a congruent $\triangle PQR$ with an opposite orientation.

Note$2$: One may observe that the ratios in $(1)$, $(2)$, $(3)$ have the form
$$k_n := \frac{sF_{n\phantom{-1}} + qF_{n+1}}{sF_{n-1}+qF_{n\phantom{+1}}}$$
where $F_n$ is the $n$-th Fibonacci number; specifically, $p=sk_1$, $r=sk_3$, $q=sk_5$.
A: You can use Heron's formula:
$$S^2=p(p-a)(p-b)(p-c)$$
where $$p=\frac12(a+b+c)$$
For the initial equilateral triangle $a=b=c$, so $$p=\frac32 a\\S^2=\frac 32\frac12\frac12\frac12 a^4=\frac 3{16}a^4$$
Due to symmetry, $S_4$ will be an equilateral triangle, and the other three triangles will be congruent.
Since $$S_1=S_2=S_3=S_4=\frac 14S$$
you get that the side of $S_4$ is $a/2$, and the sides of $S_1$ are $a$, $x$, and $x+\frac12 a$, where $x$ is the length of $AF$. Then for $S_1$ you get $$p_1=a+x+\left(\frac 12 a+x\right)=\frac 34 a+x$$
and $$S_1^2=\frac1{16}S^2\\p_1(p_1-a)(p_1-x)(p_1-x-\frac12 a)=\frac{1}{16}\frac{3}{16}a^4$$
Divide both sides by $a^4$ and you get a quadratic in $x/a$
Argument for a single solution
The above calculations rely on a symmetric solution. Let's assume that the solution is not unique. For example, one can move point $E$. Since the area $S_2$ is fixed to one quarter of $S$, it means that point $E$ move on a line parallel to $BC$. Without loss of generality, let's say that $E$ moves to the left from the symmetric case (towards the interior of the equilateral triangle). Then $\angle ECA$ increases. Since $F$ is on a line parallel to $AC$, $F$ will move closer to $C$, also towards the interior of the equilateral triangle. That means $\angle FAB$ will increase. Following the same argument as before, the new $G$ point will be inside the original equilateral triangle. Since the new $E$, $F$, $G$ are inside the original equilateral triangle, it means that the area of the new $\triangle EFG$ is smaller than $S/4$. Similarly, if we move $E$ to the right, the area of the new $\triangle EFG$ will be larger than $S/4$.
A: 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:
$x=\frac{y\pm\sqrt{y^2+4y^2}}{2}$
$\frac xy=\frac {FG}{FA}=\frac{1+\sqrt 5}2$
A: I remember doing this as a construction problem long ago, and it was an arbitrary triangle that had to be dissected, not necessarily equilateral. As it turns out, the ratio is the same in any case. You express some concern about proving the symmetry of the figure. I used dynamic geometry to come up with a couple of intuitively obvious demonstrations of that property, but found no concise proof. It is simple enough to show that a symmetric case is possible, but proof of uniqueness became rather involved.
Suppose $\triangle ABC$ and $\triangle EFG$ are concentric and both equilateral. By given conditions, $\triangle ABC$ has four times the area of $\triangle EFG$. Let $O$ be their common center. Construct apothems $OP$ and $OQ$. Let $OP = a$. By similarity of the triangles $OQ = 2a$.

Using 30°-60°-90° triangles $OFP$ and $OAQ$,
$FP = \sqrt{3}OP = \sqrt {3}a$
$OA = 2OQ = 4a$.
$AP = \sqrt {OA^2-OP^2} = \sqrt {(4a)^2-a^2} = \sqrt{15}a$
$\frac {AF}{FG} = \frac{AP-FP}{2FP} = \frac{\sqrt{15}a-\sqrt{3}a}{2\sqrt{3}a} = \frac{\sqrt{5}-1}{2}$
Edit:
As I said earlier, I found no concise proof that the interior triangle had to be equilateral. Oh well, let me see about doing so without being concise.
Here $\triangle ABC$ is drawn in black. The three blue lines are parallel to the sides of $\triangle ABC$, offset by a distance equal to one-fourth the altitude. In order to satisfy the area requirement, points $E$, $F$, and $G$ must fall on these lines.

Begin with point $E$ in motion, attached to the horizontal blue line, starting at the vertical altitude and moving to the right. Point $G$ is then constructed as the intersection of $BE$ and another blue line. As $E$ moves right, $G$ must move downward and left. Now $AG$ intersects the third blue line at $F$, which must move upward and left. In this sketch only two of the triangles satisfy the area requirement, and $\triangle CAF$ has not even been formed.
As point $E$ continues to move to the right, the length of $FG$ must increase and the distance from $E$ to $FG$ must increase. Together this means that the area of $\triangle EFG$ must be increasing continuously. The problem is solved only if the area of $\triangle EFG$ is one-fourth that of $\triangle ABC$. The continuous increase in area means that this can happen at no more than one point. Therefore, there can be no more than one solution for which point $E$ is right of center. By reflection symmetry of $\triangle ABC$, there also can be no more than one solution where $E$ is left of center.
The work I (and others) did yesterday is proof that a solution exists for an equilateral $\triangle EFG$, and that dissection could be reflected to render another having the same three-fold rotation symmetry. There can be no more than two solutions, so any solution must have that symmetry.
