Fun Equilateral Triangle Problem $\Delta ABC$ is an equilateral triangle with a side length of $4$ units.$\: $$\: $
$\angle CAF = \angle EBC =\angle FAB$ .$\: $$\: $
$D \in \left | AF \right |\: ,\: E \in \left | CD \right |\: ,\: F \in \left | BE \right | $ $\: $
Find the length of $\left | AD \right |$

By given angles, its easy to see that $\Delta DEF$ is an equilateral triangle and by similarity, $\left | EF \right |$ is $2$. Area of $\Delta ABC = 4S = 4\sqrt3 \Rightarrow S = \sqrt{3} $.
I couldn't get any further from this point
 A: Let $AD = x$. Then, $CD = 2 + x$ and $AC = 4$. Because we know $\angle ADC = \frac{2\pi}{3}$, we can apply the Law of Cosines on $\triangle ACD$:
$$AC^{2} = AD^{2} + CD^{2} - 2(AD)(CD)\cos\bigg(\frac{2\pi}{3}\bigg)$$
$$4^{2} = x^{2} + (2 + x)^{2} + x(2 + x)$$
$$3x^{2}+6x - 12 = 0$$
$$x^{2} + 2x - 4 = 0$$
The solutions are $x = -1\pm\sqrt{5}$. Taking the positive solution, we find $\boxed{AD = \sqrt{5}-1}$
A: Clearly the inner equilateral triangle with one-fourth the total area has sides of length 2.  So each outer triangle has sides measuring $x$ and $x+2$ where $x$ is the length asked for in the problem, and an included angle measuring 120°.
Now apply the side-angle-side formula for the area of a triangle:
$(1/2)(x)(x+2)\sin(120°)=\sqrt{3}$
So with $\sin(120°)=\sqrt3/2$ we get $x(x+2)=4$ from which (taking the positive root) $\color{blue}{x=\sqrt5-1}$.
The three interior segments divide one another in what was originally called "extreme and mean ratio".   The term "golden ratio" for $(1+\sqrt5)/2$ seems to have been a Renaissance invention.
Incidentally, the angle $\theta$ is close to but not exactly 45°.  More exactly it's about 44°28'.
A: No trigonometry:
Draw $FC$.  Then $$\frac{|\triangle DEF|}{|\triangle CEF|} = \frac{DE}{CE}$$ and $$\frac{|\triangle CEF|}{|\triangle BCF|} = \frac{EF}{BF} = \frac{DE}{CE}.$$  So if $|\triangle CEF| = T$, we have $$\frac{S}{T} = \frac{T}{S-T} = \frac{1}{(S/T)-1},$$ hence $$\frac{FD}{AD} = \frac{DE}{CE} = \frac{S}{T} = \frac{1 + \sqrt{5}}{2} = \varphi,$$ the golden ratio.  But since $|\triangle ABC| = 4|\triangle DEF| = 4S$, it follows that $FD = \frac{AB}{2} = 2$, hence $$AD = \frac{2}{\varphi} = \sqrt{5}-1.$$
