Inscribing Circle in Regular Pentagon
Theorem
In any given regular pentagon it is possible to inscribe a circle.
In the words of Euclid:
- In a given pentagon, which is equilateral and equiangular, to inscribe a circle.
(The Elements: Book $\text{IV}$: Proposition $13$)
Construction
Let $ABCDE$ be the given regular pentagon.
Let $\angle BCD$ and $\angle CDE$ be bisected by $CM$ and $DG$ respectively, and let these cross at $F$.
Then construct the circle whose center is $F$ and whose radius is $FM$ (or $FG$).
This circle is the one required.
Proof
Join the straight lines $FB, FA, FE$.
We have that $BC = CD$, $CF$ is common and $\angle BCF = \angle DCF$.
So by Triangle Side-Angle-Side Equality $\triangle BCF = \triangle DCF$ and so $BF = DF$.
Thus $\angle CBF = \angle CDF$.
Since $\angle CDE = 2 \angle CDF$ and $\angle CDE = \angle CBF$, then $\angle CDF = \angle CBF$.
So $\angle ABF = \angle FBC$ and so $\angle ABC$ has been bisected by the straight line $BF$.
Similarly it can be shown that $\angle BAE, \angle AED$ have been bisected by the straight lines $FA, FE$ respectively.
Now join $FH, FK, FL$ from $F$ perpendicular to $BC, CD, DE$.
We have that:
- $\angle HCF = \angle KCF$
- $\angle FHC = \angle FKC$ (both are right angles)
- $FC$ is common and subtends one of the equal angles.
So from Triangle Side-Angle-Angle Equality:
- $\triangle FHC = \triangle FKC$
and so:
- $FH = FK$
Similarly it is shown that $FL = FM = FG = FH = FK$.
Therefore the circle whose center is $F$ and radius is $FM$ (or $FG$) passes through all of the points $G, H, K, L, M$.
We have that the angles at each of those points is a right angle.
So from Line at Right Angles to Diameter of Circle, the circle $GHKLM$ is tangent to each of lines $AB, BC, CD, DE, EA$.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $13$ 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