Proof: By Euclid

Let $AC$ have been drawn at right angles to the straight line $AB$ from the point $A$ on it [Prop. 1.11], and let $AD$ have been made equal to $AB$ [Prop. 1.3].
And let $DE$ have been drawn through point $D$ parallel to $AB$ [Prop. 1.31], and let $BE$ have been drawn through point $B$ parallel to $AD$ [Prop. 1.31].
Thus, $ADEB$ is a parallelogram.
Therefore, $AB$ is equal to $DE$, and $AD$ to $BE$ [Prop. 1.34].
But, $AB$ is equal to $AD$.
Thus, the four (sides) $BA$, $AD$, $DE$, and $EB$ are equal to one another.
Thus, the parallelogram $ADEB$ is equilateral.
So I say that (it is) also right-angled.
For since the straight line $AD$ falls across the parallels $AB$ and $DE$, the (sum of the) angles $BAD$ and $ADE$ is equal to two right angles [Prop. 1.29].
But $BAD$ (is a) right angle.
Thus, $ADE$ (is) also a right angle.
And for parallelogrammic figures, the opposite sides and angles are equal to one another [Prop. 1.34].
Thus, each of the opposite angles $ABE$ and $BED$ (are) also right angles.
Thus, $ADEB$ is right-angled.
And it was also shown (to be) equilateral.
Thus, ($ADEB$) is a square [Def. 1.22] .
And it is described on the straight line $AB$.
(Which is) the very thing it was required to do.
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