Geometric proof of $\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$ It is well-known that $$\sin{20^\circ}\sin{40^\circ}\sin{80^\circ}=\frac{\sqrt{3}}{8}$$
It follows that $$\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$$
But how to prove this by geometry?
Thank you.
 A: By using Briggs formulas,
$$\sin A \sin B = \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right),$$
$$\sin C \cos D = \frac{1}{2}\left(\sin(C+D)+\sin(C-D)\right),$$
we have:
$$\begin{eqnarray*} \sin A \sin B \sin C &=& \frac{1}{2}\left(\cos(A-B)\sin C-\cos(A+B)\sin C\right)\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)+\sin(A+B-C)-\sin(A+B+C)\right)\end{eqnarray*} $$
and we just need to prove:
$$ -\sin 20^\circ + \sin 100^\circ + \sin 60^\circ -\sin 140^\circ = \sin 60^\circ $$
or:
$$ \sin 40^\circ = \sin 80^\circ- \sin 20^\circ = 2\sin 30^\circ \cos 50^\circ$$
that is trivial since $2\sin 30^\circ=1$ and $\sin 40^\circ=\cos 50^\circ$.

Another proof. The given identity is equivalent to:
$$\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{3\pi}{9}\sin\frac{4\pi}{9}=\frac{3}{16}$$
or to:
$$ \prod_{k=1}^{8}\sin\frac{k\pi}{9} = \frac{9}{256} $$
that follows from the well-known identity:

$$ \prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}.$$

A: Another Trigonometric & algebraic proof:
If $\sin3x=\sin3A,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x+\sin3A=0$
$\implies\prod_{r=1}^3\sin x_r=-\dfrac{\sin3A}4$
Again $\sin3x=\sin3A\implies3x=n180^\circ+(-1)^n3A$ where $n$ is any integer
$\implies x=n60^\circ+(-1)^nA$ 
If we choose even $n=2m,x=120^\circ m+A$
$\implies\sin A\sin(120^\circ+A)\sin(240^\circ+A)=-\dfrac{\sin3A}4$
Now use $\sin(180^\circ-y)=\sin y$ to find $\sin(120^\circ+A)=\sin[180^\circ-(60^\circ-A)]=\sin(60^\circ-A)$
and 
use $\sin(180^\circ+y)=-\sin y$ to find $\sin(240^\circ+A)=\sin[180^\circ+(60^\circ+A)]=-\sin(60^\circ+A)$
Here $A=20^\circ$
You can also choose odd $n$ and use the last formula to derive the same identity
