$\Delta ABC$ 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
Best Answer
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 $\boxed{AD = \sqrt{5}-1}$