Proof: By Euclid
(related to Proposition: 4.06: Inscribing Square in Circle)
 Let two diameters of circle $ABCD$, $AC$ and $BD$, have been drawn at right angles to one another.^{1} And let $AB$, $BC$, $CD$, and $DA$ have been joined.
 And since $BE$ is equal to $ED$, for $E$ (is) the center (of the circle), and $EA$ is common and at right angles, the base $AB$ is thus equal to the base $AD$ [Prop. 1.4].
 So, for the same (reasons), each of $BC$ and $CD$ is equal to each of $AB$ and $AD$.
 Thus, the quadrilateral $ABCD$ is equilateral.
 So I say that (it is) also rightangled.
 For since the straight line $BD$ is a diameter of circle $ABCD$, $BAD$ is thus a semicircle.
 Thus, angle $BAD$ (is) a right angle [Prop. 3.31].
 So, for the same (reasons), (angles) $ABC$, $BCD$, and $CDA$ are also each right angles.
 Thus, the quadrilateral $ABCD$ is rightangled.
 And it was also shown (to be) equilateral.
 Thus, it is a square [Def. 1.22] .
 And it has been inscribed in circle $ABCD$.
 Thus, the square $ABCD$ has been inscribed in 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