MHB Prove $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$ in Geometry Challenge

[M R]

If ABCDEFG is a regular heptagon prove that $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$.

 

[magneto1]

Spoiler

A regular polygon is cyclic. So a quadrilateral defined by any four points of the heptagon is also cyclic. Ptolemy's theorem states that the sum of the products
of two pairs of opposite equals the product of the diagonal for a cyclic quadrilateral.

Let $AB = CD = DE = x$ since they are sides of a regular polygon.
We know that $AC = CE = y$ and $AD = AE = z$. Now consider
the quadrilateral $ACDE$, and apply Ptolemy's theorem:

$$AC \cdot DE + CD \cdot AE = AD \cdot CE,$$ or using the notation, this becomes: $y x + x z = z y$. Or $x(y + z) = zy$. This further simplify to:
$\frac 1x = \frac{y + z}{zy} = \frac 1y + \frac 1z$. Therefore,
$\frac 1{AB} = \frac 1{AC} + \frac 1{AD}$

 

[M R]

Spoiler

A regular polygon is cyclic. So a quadrilateral defined by any four points of the heptagon is also cyclic. Ptolemy's theorem states that the sum of the products
of two pairs of opposite equals the product of the diagonal for a cyclic quadrilateral.

Let $AB = CD = DE = x$ since they are sides of a regular polygon.
We know that $AC = CE = y$ and $AD = AE = z$. Now consider
the quadrilateral $ACDE$, and apply Ptolemy's theorem:

$$AC \cdot DE + CD \cdot AE = AD \cdot CE,$$ or using the notation, this becomes: $y x + x z = z y$. Or $x(y + z) = zy$. This further simplify to:
$\frac 1x = \frac{y + z}{zy} = \frac 1y + \frac 1z$. Therefore,
$\frac 1{AB} = \frac 1{AC} + \frac 1{AD}$

Neatly done. :)