Inscribing Circle in Square

Construction

 * Euclid-IV-8.png

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.