Circumscribing Square about Circle
Theorem
In the words of Euclid:
- About a given circle to circumscribe a square.
(The Elements: Book $\text{IV}$: Proposition $7$)
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.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{IV}$. Propositions