Inscribing Regular Pentagon in Circle

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

Construction

 * Euclid-IV-11.png

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.