Proof: By Euclid
(related to Proposition: 4.08: Inscribing Circle in Square)
 Let $AD$ and $AB$ each have been cut in half at points $E$ and $F$ (respectively) [Prop. 1.10].
 And let $EH$ have been drawn through $E$, parallel to either of $AB$ or $CD$, and let $FK$ have been drawn through $F$, parallel to either of $AD$ or $BC$ [Prop. 1.31].
 Thus, $AK$, $KB$, $AH$, $HD$, $AG$, $GC$, $BG$, and $GD$ are each parallelograms, and their opposite sides [are] manifestly equal [Prop. 1.34].
 And since $AD$ is equal to $AB$, and $AE$ is half of $AD$, and $AF$ half of $AB$, $AE$ (is) thus also equal to $AF$.
 So that the opposite (sides are) also (equal).
 Thus, $FG$ (is) also equal to $GE$.
 So, similarly, we can also show that each of $GH$ and $GK$ is equal to each of $FG$ and $GE$.
 Thus, the four (straight lines) $GE$, $GF$, $GH$, and $GK$ [are] equal to one another.
 Thus, the circle drawn with center $G$, and radius one of $E$, $F$, $H$, or $K$, will also go through the remaining points.
 And it will touch the straight lines $AB$, $BC$, $CD$, and $DA$, on account of the angles at $E$, $F$, $H$, and $K$ being right angles.
 For if the circle cuts $AB$, $BC$, $CD$, or $DA$, then a (straight line) drawn at right angles to a diameter of the circle, from its extremity, will fall inside the circle.
 The very thing was shown (to be) absurd [Prop. 3.16].
 Thus, the circle drawn with center $G$, and radius one of $E$, $F$, $H$, or $K$, does not cut the straight lines $AB$, $BC$, $CD$, or $DA$.
 Thus, it will touch them, and will have been inscribed in the square $ABCD$.
 Thus, a circle has been inscribed in the given square.
 (Which is) the very thing it was required to do.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"