Inscribing Circle in Square
Theorem
In the words of Euclid:
(The Elements: Book $\text{IV}$: Proposition $8$)
Construction
Let $\Box ABCD$ be the given square.
Let $AD, AB$ be bisected at $E$ and $F$ respectively.
Let $EH$ be drawn parallel to $AB$ or $CD$.
Let $FK$ be drawn parallel to $AD$ or $BC$.
Let $G$ be where $EH$ crosses $FK$.
Draw the circle $EFHK$ whose center is at $G$ and whose radius is $EG$.
Then $EFHK$ is the required circle.
Proof
Each of the figures $AK, KB, AH, FD, AG, GC, BG, GD$ is a parallelogram from the method of construction.
The opposite sides of all of them is equal.
We have that $AD = AB$ and $2 AE = AD$, and $2 AF = AB$.
Therefore $AE = AF$.
Because the opposite sides are equal, $FG = GE$.
Similarly we can show that $GE = GF = GH = GK$.
So the circle whose center is at $G$ and whose radius is $EG$ passes through all of $E, F, H, K$.
The lines $AB, BC, CD, DA$ are all perpendicular to the diameters $FK$ and $EH$.
So from Line at Right Angles to Diameter of Circle, these lines are tangent to the circle.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $8$ 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