Inscribing Circle in Regular Pentagon

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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

Euclid-IV-13.png

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 theorem is Proposition $13$ of Book $\text{IV}$ of Euclid's The Elements.


Sources