Circumscribing Regular Pentagon about Circle

Theorem

About a given circle, it is possible to circumscribe a regular pentagon.

In the words of Euclid:

About a given circle to circumscribe an equilateral and equiangular pentagon.

(The Elements: Book $\text{IV}$: Proposition $12$)

Construction

Let $ABCDE$ be the given circle.

Find the center $F$ of the circle $ABCDE$.

Let $A, B, C, D, E$ be the vertices of a regular pentagon which has been inscribed within the circle $ABCDE$.

Draw tangents $GH, HK, KL, LM, MG$ to $ABCDE$ at the points $A, B, C, D, E$.

Then $GHKLM$ is the required regular pentagon.

Proof

Join $FB, FK, FC, FL, FD$.

From Radius at Right Angle to Tangent, $HK, KL$ etc. are perpendicular to the radii they touch.

So $\angle KCF$ and $\angle LCF$ are right angles.

For the same reason, $\angle KBF$ and $\angle LDF$ are right angles.

By Pythagoras's Theorem, $FK^2 = FC^2 + CK^2$

For the same reason, $FK^2 = FB^2 + BK^2$.

As $FB = FC$ it follows that $BK = CK$.

From Triangle Side-Side-Side Congruence it follows that $\triangle FCK = \triangle FBK$ and so $\angle FKC = \angle FKB$ and $\angle KFC = \angle KFB$.

So $\angle BFC = 2 \angle KFC$ and $\angle BKC = 2 \angle FKC$.

For the same reason $\angle CFD = 2 \angle CFL$ and $\angle DLC = 2 \angle FLC$.

We have that the arc $BC$ equals the arc $CD$.

So from Angles on Equal Arcs are Equal, $\angle BFC = \angle CFD$.

As $\angle BFC = 2 \angle KFC$ and $\angle CFD = 2 \angle CFL$, it follows that $\angle KFC = \angle CFL$.

But $\angle FCK = \angle FCL$.

So from Triangle Angle-Side-Angle Congruence:

$\triangle FCK = \triangle FLC$

So $KC = CL$ and $\angle FKC = \angle FLC$.

Since $KC = CL$ it follows that $KL = 2 KC$.

For the same reason $HK = 2 BK$ and $BK = KC$.

So $HK = KL$.

Similarly each of the straight lines $HG, GM, ML$ are all equal to $HK$ and $KL$.

So the pentagon $GHKLM$ is equilateral.

We have that:

$\angle FKC = \angle FLC$

$\angle HKL = 2 \angle FKC$

$\angle KLM = 2 \angle FLC$

So:

$\angle HKL = \angle KLM$

Similarly each of $\angle KHG, \angle HGM, \angle GML$ are equal to $\angle HKL$ and $\angle KLM$.

So the pentagon $GHKLM$ is equiangular.

$\blacksquare$

Historical Note

This proof is Proposition $12$ 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