Gauss sum in regular 7-gon 
The heptagon on the picture is a regular heptagon with side 1. What is the length of the dashed interval?


This is a (kind of) 'geometric version of a quadratic Gauss sum for p=7' (this observation — and the problem — is due to Dušan Djukić, I believe): once you know that $\sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{8\pi}7=\frac{\sqrt7}2$ finding the answer is easy. But I'm interested in going in the opposite direction — so I'm asking for a geometric proof (not using trigonometry / complex numbers).

It would also be nice to have a generalisation to 11-gon (and beyond). One possible geometric interpretation of quadratic Gauss sum for p=11 is on the picture below (but maybe there is a better choice…).

 A: Let $P_0 P_1 P_2 P_3 P_4 P_5 P_6$ be a regular heptagon with side-length $p$ and $\theta := \angle P_5 P_1 P_2 = \angle P_6 P_2 P_3$. Let $\overline{P_1 P_5}$ meet $\overline{P_2 P_6}$ at $Q$. Define $q :=|\overline{QP_2}| = |\overline{QP_5}|$ and $d := |\overline{QP_3}|$. 


Claim. $\quad d = p\sqrt{2}$

Because $\overline{P_1 P_5}\parallel\overline{P_0P_6}$ and $\overline{P_2 P_6}\parallel\overline{P_1P_0}$, we have that $\square P_0 P_1 Q P_6$ is a parallelogram; in particular, it is a rhombus, so that $|\overline{QP_1}| = p$.
Using a pinch of trig in the form of the Law of Cosines, we observe
$$\begin{align}
\triangle QP_2 P_3: \quad d^2 &= p^2 + q^2 - 2 pq \cos\theta \tag{1a}\\
\triangle QP_1 P_2: \quad q^2 &= 2p^2-2p^2\cos\theta \tag{1b}
\end{align}$$
Thus, replacing $q^2$ in $(1a)$ using $(1b)$,
$$d^2 = 3 p^2 - 2p(p+q)\cos\theta \tag{2}$$
However, writing $M$ for the midpoint of $\overline{P_1 P_2}$, we see in $\triangle P_5 P_1 M$ that
$$(p+q)\cos\theta = \frac{1}{2}p \tag{3}$$
which implies
$$d^2 = 2 p^2 \tag{4}$$
and proves the Claim. $\square$
