Inscribing Regular Hexagon in Circle

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

Construction

 * Euclid-IV-15.png

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.