Construction of Square on Given Straight Line

Theorem
On any given line segment, it is possible to construct a square.

Proof

 * Euclid-I-46.png

Let $$AB$$ be the given line segment.

Construct$AC$ perpendicular to $$AB$$.

By Construction of Equal Straight Lines from Unequal, place $$D$$ on $$AC$$ so $$AD = AB$$.

Construct $DE$ parallel to $AB$, and construct $BE$ parallel to $AD$.

So $$ADEB$$ is a parallelogram.

So $$AB = DE$$ and $$AD = BE$$ from Opposite Sides and Angles of Parallelogram are Equal.

So parallelogram is equilateral.

Also, as $$AD$$ falls on the parallels $$AB$$ and $$DE$$, from Parallel Implies Supplementary Interior Angles we have that $$\angle BAD + \angle ADE$$ equals two right angles.

But as $$\angle BAD$$ is right, so is $$\angle ADE$$.

And, from Opposite Sides and Angles of Parallelogram are Equal, so are $$\angle ABE$$ and $$\angle BED$$.

So $$ADEB$$ is equilateral and equiangular, and therefore, by definition, a square.