Inscribing Square in Circle

Theorem

In the words of Euclid:

In a given circle to inscribe a square.

(The Elements: Book $\text{IV}$: Proposition $6$)

Construction

Let $ABCD$ be the given circle.

Let two diameters be drawn at right angles to one another.

Join $AB, BC, CD, DA$.

Then $\Box ABCD$ is the required square.

Proof

We have that $BE = ED$, $EA$ is common and $\angle BEA = \angle DEA$ are right angles.

So from Triangle Side-Angle-Side Congruence, $\triangle ABE = \triangle ADE$ and so $AB = AD$.

For the same reason $BC = CD = AD$ and so all four sides $AB, BC, CD, DA$ are equal.

So $\Box ABCD$ is equilateral.

Next we have that $BD$ is a diameter of circle $ABCD$.

So from Relative Sizes of Angles in Segments, $\angle BAD$ is a right angle.

For the same reason, $\angle ABC$, $\angle BCD$ and $\angle ADC$ are also all right angles.

So by definition, $\Box ABCD$ is a square.

$\blacksquare$

Historical Note

This proof is Proposition $6$ 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.)