Proving correctness of a braced regular heptagon from a trigonometric identity This is a rigid regular heptagon I found on Wikipedia during associated research for my question on rigid pentagons:

The accompanying text reads

The construction includes two isosceles triangles which hold the rest of [the] bars fixed. The regular heptagon's side $a$, the shorter isosceles triangle side $e$, and the longer isosceles triangle side $d$ satisfy
  $$7a^2+e^2=4d^2\tag1$$

Here $a,d,e$ are integral, or more generally rational. The Wikipedia text goes on to say that $(1)$ can be derived from the following identity:
$$\sin\frac\pi7-\sin\frac{2\pi}7-\sin\frac{4\pi}7=-\frac{\sqrt7}2\tag2$$
(If $a=1$ then by solving a Pell equation (see e.g. here) we get $d=\frac t{16}+\frac7t$ for $t\in\mathbb Q$.)
Now it is easy to prove $(2)$ by a minimal polynomial calculation. It is also easy to prove $(1)$ once you determine that the isosceles triangle's height is $\frac{\sqrt7}2a$. But how does $(1)$ follow from $(2)$?
If the association can be shown, I might be able to derive a rigid regular nonagon with rational sticks from the following identity.
$$\sin\frac\pi9+\sin\frac{2\pi}9=\sin\frac{4\pi}9$$
 A: Rewrite $\sin\frac\pi7$ as $-\sin\frac{8\pi}7$ and $(1)$ as
$$\sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{8\pi}7=\frac{\sqrt7}2$$
But this is also easy to show:
$$\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7=-\frac12$$
Now suppose the heptagon has unit side length and one side spans $-1$ and $0$ (we are treating points as equivalent to complex numbers here). Erect the isosceles triangle discussed in the Wikipedia text on that side, pointing upwards. In a very natural way induced by the construction, the apex of that triangle is $e^{2\pi/7}+e^{4\pi/7}+e^{8\pi/7}$, which by the two identities above equals $\frac12(-1+\sqrt7i)$. Thus the height of the isosceles triangle is $\frac{\sqrt7}2$ and $(2)$ follows.
A: The same argument yet in an other fashion... (i started to draw and in the meantime there were already two solutions...)

In the picture we have
$$
\begin{aligned}
d^2-\frac 14e^2 &= E'I^2-JI^2
\\
&=E'J^2=(JL+LK-KE')^2\\
&=\left(
BC\sin\frac {2\pi}7
+
CD\sin\frac{3\pi}7
-
DE'\sin\frac\pi7
\right)^2
\\
&=\left(\frac {a\sqrt 7}2\right)^2
=\frac 74a^2\ .
\end{aligned}
$$
A: The height of the regular heptagon, from the base of an isosceles triangle to the opposite vertex, is $$a\left(\sin\frac{\pi}7 +\sin\frac{2\pi}7+\sin\frac{3\pi}7\right)$$
The shorter diagonal of a rhombus is $2a\sin\frac\pi7$ so the altitude of the isosceles triangle is 
$$a\left(-\sin\frac{\pi}7 +\sin\frac{2\pi}7+\sin\frac{3\pi}7\right)=\sqrt7a/2$$
Half the base of the isosceles triangle is $e/2$, then pythagoras gives (1)
