Proof: By Euclid

$AC$ and $BD$ being joined, let them cut one another at $E$.
And since $DA$ is equal to $AB$, and $AC$ (is) common, the two (straight lines) $DA$, $AC$ are thus equal to the two (straight lines) $BA$, $AC$.
And the base $DC$ (is) equal to the base $BC$.
Thus, angle $DAC$ is equal to angle $BAC$ [Prop. 1.8].
Thus, the angle $DAB$ has been cut in half by $AC$.
So, similarly, we can show that $ABC$, $BCD$, and $CDA$ have each been cut in half by the straight lines $AC$ and $DB$.
And since angle $DAB$ is equal to $ABC$, and $EAB$ is half of $DAB$, and $EBA$ half of $ABC$, $EAB$ is thus also equal to $EBA$.
So that side $EA$ is also equal to $EB$ [Prop. 1.6].
So, similarly, we can show that each of the [straight lines] $EA$ and $EB$ are also equal to each of $EC$ and $ED$.
Thus, the four (straight lines) $EA$, $EB$, $EC$, and $ED$ are equal to one another.
Thus, the circle drawn with center $E$, and radius one of $A$, $B$, $C$, or $D$, will also go through the remaining points, and will have been circumscribed about the square $ABCD$.
Let it have been (so) circumscribed, like $ABCD$ (in the figure).
Thus, a circle has been circumscribed about the given square.
(Which is) the very thing it was required to do.
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