Proof: By Euclid

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 right-angled.
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 right-angled.
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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Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right angles to the first [Prop. 1.11] (translator's note). ↩