# Inscribing Regular Hexagon in Circle

## Theorem

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

In the words of Euclid:

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

(*The Elements*: Book $\text{IV}$: Proposition $15$)

### Porism

In the words of Euclid:

*From this it is manifest that the side of the hexagon is equal to the radius of the circle.*

And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon.

And further by means to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it.

(*The Elements*: Book $\text{IV}$: Proposition $15$ : Porism)

## Construction

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

Let $AD$ be a diameter of the circle $ABCDEF$.

Let the center $G$ be found.

Draw the circle $EGCH$ with center $D$ and radius $DG$.

Join $EC, CG$ and produce them to $B$ and $F$ on the circumference of the circle $ABCDEF$.

Join $AB, BC, CD, DE, EF, FA$.

This is the required regular hexagon.

## Proof

Since $G$ is the center of circle $ABCDEF$, it follows that $GE = GD$.

Since $D$ is the center of circle $EGCH$, it follows that $DE = GD$.

So $GE = GD = DE$ and so $\triangle EGD$ is equilateral and so equiangular.

By Sum of Angles of Triangle Equals Two Right Angles, $\angle EGD$ is one third of two right angles.

Similarly for $\angle DGC$.

Since the straight line $CG$ on $EB$ makes $\angle EGC + \angle CGB$ equal to two right angles, $\angle CGB$ is also equal to one third of two right angles.

So $\angle EGD = \angle DGC = \angle CGB$.

By the Vertical Angle Theorem, $\angle BGA = \angle AGF = \angle FGE = \angle EGD = \angle DGC = \angle CGB$.

From Equal Angles in Equal Circles, the six arcs $AB, BC, CD, DE, EF, FA$ are all equal.

So the six straight lines $AB, BC, CD, DE, EF, FA$ are all equal.

So the hexagon $ABCDEF$ is equilateral.

Now since the arc $FA$ equals the arc $ED$, let the arc $ABCD$ be added to each.

So arc $FABCD$ equals arc $ABCDE$.

Now $\angle FED$ stands on arc $FABCD$ and $\angle AFE$ stands on arc $ABCDE$.

So by Angles on Equal Arcs are Equal $\angle FED = \angle AFE$.

In the same way we show that all the angles around the hexagon $ABCDEF$ are equal.

Therefore $ABCDEF$ is a regular hexagon.

$\blacksquare$

## Historical Note

This proof is Proposition $15$ 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 - 1989: Ephraim J. Borowski and Jonathan M. Borwein:
*Dictionary of Mathematics*... (previous) ... (next):**hexagon** - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)