On a given straight line to describe a square.

Let $AB$ be the given straight line;thus it is required to describe a square on the straight line $AB$.Let $AC$ be drawn at right angles to the straight line $AB$ from the point $A$ on it [Prop. 1.11], and let $AD$ be made equal to $AB$;through the point $D$ let $DE$ be drawn parallel to $AB$,and through the point $B$ let $BE$ be drawn parallel to $AD$. [Prop. 1.31]Therefore $ADEB$ is a parallelogram;therefore $AB$ is equal to $DE$, and $AD$ to $BE$. [Prop. 1.34]But $AB$ is equal to $AD$;therefore the four straight lines $BA$, $AD$, $DE$, $EB$ are equal to one another;therefore the parallelogram $ADEB$ is equilateral.I say next that it is also right-angled.For, since the straight line $AD$ falls upon the parallels $AB$, $DE$,the angles $BAD$, $ADE$ are equal to two right angles. [Prop. 1.29]But the angle $BAD$ is right;therefore the angle $ADE$ is also right.And in parallelogrammic areas the opposite sides and angles are equal to one another; [Prop. 1.34]threfore each of the opposite angles $ABE$, $BED$ is also right.Therefore $ADEB$ is right-angled.And it was also proved equilateral.Therefore it is a square; and it is described on the straight line $AB$.Q.E.F.