# Inscribing Regular Pentagon in Circle

## Theorem

In a given circle, it is possible to inscribe a regular pentagon.

In the words of Euclid:

*In a given circle to inscribe an equilateral and equiangular pentagon.*

(*The Elements*: Book $\text{IV}$: Proposition $11$)

## Construction

Let $ABCDE$ be the given circle (although note that at this stage the positions of the points $A, B, C, D, E$ have not been established).

Let $\triangle FGH$ be constructed such that $\angle FGH = \angle FHG = 2 \angle GFH$.

Let $ACD$ be inscribed in $ABCDE$ such that $\angle ACD = \angle FGH, \angle ADC = \angle FHG, \angle CAD = \angle GFH$.

Bisect $\angle ACD$ with $CE$ and bisect $\angle ADC$ with $DB$.

Then the pentagon $ABCDE$ is the required regular pentagon.

## Proof

We have that $\angle CDA = \angle DCA = 2 \angle CAD$.

As $\angle CDA$ and $ \angle DCA$ have been bisected, $\angle DAC = \angle ACE = \angle ECD = \angle CDB = \angle BDA$.

From Equal Angles in Equal Circles it follows that the arcs $AB, BC, CD, DE, EA$ are all equal.

Hence from Equal Arcs of Circles Subtended by Equal Straight Lines, the straight lines $AB, BC, CD, DE, EA$ are all equal.

So the pentagon $ABCDE$ is equilateral.

Now since the arc $AB$ equals arc $DE$, we can add $BCD$ to each.

So the arc $ABCD$ equals arc $BCDE$.

So from Angles on Equal Arcs are Equal $\angle BAE = \angle AED$.

For the same reason, $\angle BAE = \angle AED = \angle ABC = \angle BCD = \angle CDE$.

So the pentagon $ABCDE$ is equiangular.

$\blacksquare$

## Historical Note

This proof is Proposition $11$ of Book $\text{IV}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{IV}$. Propositions - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)