Circumscribing a Circle about a Square

Theorem

About any given square it is possible to circumscribe a circle.

Construction

Let $\Box ABCD$ be the given square.

Join $AC$ and $BD$ and let their point of intersection be $E$.

Draw the circle whose center is $E$ and whose radius is $AE$.

The circle $ABCD$ is the required circle.

Proof

We have that $DA = AB$, $DC = BC$ and $AC$ is common.

So from Triangle Side-Side-Side Equality $\triangle DAB = \triangle DCB$ and so $\angle DAC = \angle BAC$.

So $\angle DAB$ is bisected by $AC$.

Similarly each of $\angle ABC, \angle BCD, \angle DCA$ are bisected by $AC$ or $DB$.

We have that $\angle DAB = \angle ABC$, $2 \angle EAB = \angle DAB$, $2 \angle EBA = \angle ABC$.

Then $\angle EAB = \angle EBA$.

From Triangle with Two Equal Angles is Isosceles, $EA = EB$.

Similarly, $EA = EB = EC = ED$.

So the circle whose center is $E$ and whose radius is $AE$ passes through each of $B, C, D$.

Hence the result.

$\blacksquare$

Historical Note

This is Proposition 9 of Book IV of Euclid's The Elements.