Perpendicular from Center of Circle to Side of Inscribed Pentagon
Theorem
In the words of Hypsicles of Alexandria:
- The perpendicular drawn from the centre of any circle to the side of the pentagon inscribed in the same circle is half the sum of the side of the hexagon and the decagon inscribed in the same circle.
(The Elements: Book $\text{XIV}$: Proposition $1$)
Proof
Let $ABC$ be a circle.
Let $BC$ be the side of a regular pentagon which has been inscribed within $ABC$.
Let $D$ be the center of the circle $ABC$.
Let $DE$ be drawn from $D$ perpendicular to $BC$, and produce it both ways to meet $ABC$ in $A$ and $F$.
It is to be demonstrated that $DE$ is half the sum of:
- the side of a regular hexagon which has been inscribed within $ABC$
and:
- the side of a regular decagon which has been inscribed within $ABC$
Let $DC$ and $CF$ be joined.
Let $G$ be the point on $AE$ such that $GE = EF$.
Let $GC$ be joined.
We have that:
- the circumference of $ABC$ is $5$ times $\smile BFC$
and:
- half the circumference of $ABC$ is $\smile ACF$
where $\smile BFC$ and $\smile ACF$ are used to denote the arc $BFC$ and arc $ACF$.
while:
- $\smile FC$ is half of $\smile BFC$.
Thus:
| \(\ds \smile ACF\) | \(=\) | \(\ds 5 \cdot \smile FC\) | ||||||||||||
| \(\ds \therefore \ \ \) | \(\ds \smile AC\) | \(=\) | \(\ds 4 \cdot \smile CF\) | |||||||||||
| \(\ds \therefore \ \ \) | \(\ds \angle ADC\) | \(=\) | \(\ds 4 \cdot \angle CDF\) | |||||||||||
| \(\ds \therefore \ \ \) | \(\ds \angle AFC\) | \(=\) | \(\ds 2 \cdot \angle CDF\) | |||||||||||
| \(\ds \therefore \ \ \) | \(\ds \angle GCF\) | \(=\) | \(\ds \angle AFC\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cdot \angle CDF\) | ||||||||||||
| \(\ds \therefore \ \ \) | \(\ds \angle CDG\) | \(=\) | \(\ds \angle DCG\) | Sum of Angles of Triangle equals Two Right Angles | ||||||||||
| \(\ds \therefore \ \ \) | \(\ds DG\) | \(=\) | \(\ds CG\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds CF\) |
Also:
| \(\ds GE\) | \(=\) | \(\ds EF\) | ||||||||||||
| \(\ds \therefore \ \ \) | \(\ds DE\) | \(=\) | \(\ds EF + FC\) | |||||||||||
| \(\ds \therefore \ \ \) | \(\ds 2 \cdot DE\) | \(=\) | \(\ds DF + FC\) | adding $DE$ to each |
But from Porism to Proposition $15$ of Book $\text{VI} $: Inscribing Regular Hexagon in Circle:
- $DE$ is the side of a regular hexagon which has been inscribed within $ABC$
and from the construction:
- $FC$ is the side of a regular decagon which has been inscribed within $ABC$.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $1$ of Book $\text{XIV}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): The So-Called Book $\text{XIV}$, by Hypsicles