compass and straightedge construction of square
One can construct a square with sides of a given length $s$ using compass and straightedge as follows:

1.
Draw a line segment^{} of length s. Label its endpoints^{} $P$ and $Q$.

2.
Extend the line segment past $Q$.

3.
Erect the perpendicular^{} to $\overrightarrow{PQ}$ at $Q$.

4.
Using the line drawn in the previous step, mark off a line segment of length $s$ such that one of its endpoints is $Q$. Label the other endpoint as $R$.

5.
Draw an arc of the circle with center $P$ and radius $\overline{PQ}$.

6.
Draw an arc of the circle with center $R$ and radius $\overline{QR}$ to find the point $S$ where it intersects the arc from the previous step such that $S\ne Q$.

7.
Draw the square $PQRS$.
This construction is justified because $PS=PQ=QR=QS$, yielding that $PQRS$ is a rhombus^{}. Since $\mathrm{\angle}PQR$ is a right angle^{}, it follows that $PQRS$ is a square.
If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.
Title  compass and straightedge construction of square 

Canonical name  CompassAndStraightedgeConstructionOfSquare 
Date of creation  20130322 17:19:13 
Last modified on  20130322 17:19:13 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  5 
Author  Wkbj79 (1863) 
Entry type  Algorithm 
Classification  msc 51M15 
Classification  msc 5100 