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.)