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compass and straightedge construction of square
One can construct a square with sides of a given length $s$ using compass and straightedge as follows:
- Draw a line segment of length s. Label its endpoints $P$ and $Q$ .
{.} \psline[linecolor=blue](-2,0)... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img1.png)
- Extend the line segment past $Q$ .
{.} \rput[r](3.5,0){.} \psline(-2... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img2.png)
- Erect the perpendicular to $\overrightarrow{PQ}$ at $Q$ .
{.} \rput[a](2,4.9){.} \rput[r](3.5,... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img3.png)
- 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 $R$ .
{.} \rput[a](2,4.9){.} \rput[r](3.5,... ...](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img4.png)
- Draw an arc of the circle with center $P$ and radius $\overline{PQ}$ .
{.} \rput[b](2,-2){.} \rput[... ...](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img5.png)
- 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 $S \neq Q$ .
{.} \rput[b](2,-2){.} \rput[... ...a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \rput[b](-2.2,4.2){$S$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img6.png)
- Draw the square $PQRS$ .
{.} \rput[b](2,-2){.} \rput[... ...a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \rput[b](-2.2,4.2){$S$} \end{pspicture}](http://images.planetmath.org/cache/objects/9671/js/img7.png)
This construction is justified because $PS=PQ=QR=QS$ , yielding that $PQRS$ is a rhombus. Since $\angle PQR$ 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 link provided.
compass and straightedge construction of square is owned by Warren Buck.
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