Proof: By Euclid
(related to Proposition: 4.07: Circumscribing Square about Circle)
 Let two diameters of circle $ABCD$, $AC$ and $BD$, have been drawn at right angles to one another.^{1} And let $FG$, $GH$, $HK$, and $KF$ have been drawn through points $A$, $B$, $C$, and $D$ (respectively), touching circle $ABCD$.^{2}
 Therefore, since $FG$ touches circle $ABCD$, and $EA$ has been joined from the center $E$ to the point of contact $A$, the angles at $A$ are thus right angles [Prop. 3.18].
 So, for the same (reasons), the angles at points $B$, $C$, and $D$ are also right angles.
 And since angle $AEB$ is a right angle, and $EBG$ is also a right angle, $GH$ is thus parallel to $AC$ [Prop. 1.29].
 So, for the same (reasons), $AC$ is also parallel to $FK$.
 So that $GH$ is also parallel to $FK$ [Prop. 1.30].
 So, similarly, we can show that $GF$ and $HK$ are each parallel to $BED$.
 Thus, $GK$, $GC$, $AK$, $FB$, and $BK$ are (all) parallelograms.
 Thus, $GF$ is equal to $HK$, and $GH$ to $FK$ [Prop. 1.34].
 And since $AC$ is equal to $BD$, but $AC$ (is) also (equal) to each of $GH$ and $FK$, and $BD$ is equal to each of $GF$ and $HK$ [Prop. 1.34] [and each of $GH$ and $FK$ is thus equal to each of $GF$ and $HK$], the quadrilateral $FGHK$ is thus equilateral.
 So I say that (it is) also rightangled.
 For since $GBEA$ is a parallelogram, and $AEB$ is a right angle, $AGB$ is thus also a right angle [Prop. 1.34].
 So, similarly, we can show that the angles at $H$, $K$, and $F$ are also right angles.
 Thus, $FGHK$ is rightangled.
 And it was also shown (to be) equilateral.
 Thus, it is a square [Def. 1.22] .
 And it has been circumscribed about circle $ABCD$.
 Thus, a square has been circumscribed about the given circle.
 (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"
Footnotes