The regular heptagon identity

Chord lengths. A chord joining two vertices $k$ steps apart has length depending only on $\min(k,7-k)$:
- $1$ step: length $a$;
- $2$ steps: length $b$;
- $3$ steps: length $c$; and $A_5A_1$ spans $4$ steps $=3$ the other way, so $A_5A_1=c$.

Ptolemy. The quadrilateral $A_1A_2A_3A_5$ is cyclic with vertices in this order. Its sides are $a,a,b,c$; its diagonals are $b,c$. Ptolemy:
$$A_1A_3\cdot A_2A_5=A_1A_2\cdot A_3A_5+A_2A_3\cdot A_5A_1,$$
so $bc=ab+ac$. Divide by $abc$:
$$\frac1a=\frac1c+\frac1b.\ \blacksquare$$

Remark. Numerically with circumradius $1$: $a\approx0.868$, $b\approx1.564$, $c\approx1.950$, and $\frac1b+\frac1c\approx1.152=\frac1a$.

Lesson. In a regular polygon, many chords are equal; Ptolemy on a well-chosen quadrilateral gives algebraic relations between the distinct chord lengths.