## Theorem

In the words of Euclid:

About a given circle to circumscribe a square.

## Construction

Let $ABCD$ be the given circle.

Let two diameters$AC, BD$ be drawn at right angles to one another.

Through $A, B, C, D$ draw $FG, GH, HK, KF$ perpendicular to $AC, BD$.

Then $\Box FGHK$ is the required square.

## Proof

From Line at Right Angles to Diameter of Circle, $FG, GH, HK, KF$ will be tangent to $ABCD$.

Since $\angle AEB$ is a right angle, $\angle EBG$ is also a right angle.

So $GH$ is parallel to $AC$.

For the same reason $FK$ is parallel to $AC$.

From Parallelism is Transitive Relation, $GH$ is parallel to $FK$.

Similarly, each of $GF, HK$ is parallel to $BD$.

So $GK, GC, AK, FB, BK$ are all parallelograms.

Therefore from Opposite Sides and Angles of Parallelogram are Equal $GF = HK$ and $GH = FK$.

We have that $AC = BD, AC = GH, AC = FK$.

Also $BD = GF = HK$.

So $\Box FGHK$ is equilateral.

Since $GBEA$ is a parallelogram, $\angle AEB$ is a right angle.

Therefore $\angle AGB$ is also a right angle.

Similarly the angles at $H, K, F$ are also right angles.

So $\Box FGHK$ is a square.

$\blacksquare$

## Historical Note

This proof is Proposition $7$ of Book $\text{IV}$ of Euclid's The Elements.