Given line segments of lengths $a$ and $b$ , one can construct a line segment of length $\sqrt{ab}$ using compass and straightedge as follows:

1. Draw a line segment of length $a$ . Label its endpoints $A$ and $C$ .
   \begin{pspicture}(-3,-2)(3,3) \rput[r](-0.4,0){.} \psline[linecolor=blue](-2.5,0)(-0.5,0) \psdots(-2.5,0)(-0.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \end{pspicture}
2. Extend the line segment past $C$ .
   \begin{pspicture}(-3,-2)(3,3) \rput[r](3,0){.} \psline(-2.5,0)(-0.5,0) \psline[linecolor=blue]{->}(-0.5,0)(3,0) \psdots(-2.5,0)(-0.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \end{pspicture}
3. Mark off a line segment of length $b$ such that one of its endpoints is $C$ . Label its other endpoint as $B$ .
   \begin{pspicture}(-3,-2)(3,3) \rput[r](3,0){.} \psline(-2.5,0)(-0.5,0) \psline{->}(2.5,0)(3,0) \psline[linecolor=blue](-0.5,0)(2.5,0) \psdots(-2.5,0)(-0.5,0)(2.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \rput[a](2.5,-0.3){$B$} \end{pspicture}
4. Construct the perpendicular bisector of $\overline{AB}$ in order to find its midpoint $M$ .
   \begin{pspicture}(-3,-3)(3,3) \rput[r](3,0){.} \rput[a](0,3){.} \rput[b](0,-3){.} \psline{->}(-2.5,0)(3,0) \psarc[linecolor=blue](-2.5,0){3}{-45}{45} \psarc[linecolor=blue](2.5,0){3}{135}{225} \psline[linecolor=blue]{<->}(0,-3)(0,3) \psdots(-2.5,0)(-0.5,0)(0,0)(2.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \rput[a](0.3,-0.3){$M$} \rput[a](2.5,-0.3){$B$} \end{pspicture}
5. Construct a semicircle with center $M$ and radii $\overline{AM}$ and $\overline{BM}$ .
   \begin{pspicture}(-3,-3)(3,3) \rput[r](3,0){.} \rput[a](0,3){.} \rput[b](0,-3){.} \psline{->}(-2.5,0)(3,0) \psarc(-2.5,0){3}{-45}{45} \psarc(2.5,0){3}{135}{225} \psline{<->}(0,-3)(0,3) \psarc[linecolor=blue](0,0){2.5}{0}{180} \psdots(-2.5,0)(-0.5,0)(0,0)(2.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \rput[a](0.3,-0.3){$M$} \rput[a](2.5,-0.3){$B$} \end{pspicture}
6. Erect the perpendicular to $\overline{AB}$ at $C$ to find the point $D$ where it intersects the semicircle. The line segment $\overline{DC}$ is of the desired length.
   \begin{pspicture}(-3,-3)(3,3) \rput[r](3,0){.} \rput[a](0,3){.} \rput[b](0,-3){.} \psline{->}(-2.5,0)(3,0) \psarc(-2.5,0){3}{-45}{45} \psarc(2.5,0){3}{135}{225} \psline{<->}(0,-3)(0,3) \psarc(0,0){2.5}{0}{180} \psarc[linecolor=blue](-0.5,0){0.5}{160}{380} \psarc[linecolor=blue](-1,0){0.7}{-60}{60} \psarc[linecolor=blue](0,0){0.7}{120}{240} \psline[linecolor=blue]{<-}(-0.5,-1)(-0.5,2.45) \psdots(-2.5,0)(-0.5,0)(0,0)(2.5,0)(-0.5,2.45) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \rput[a](0.3,-0.3){$M$} \rput[a](2.5,-0.3){$B$} \rput[b](-0.5,2.65){$D$} \end{pspicture}

   This construction is justified because, if $\overline{AD}$ and $\overline{BD}$ were drawn, then the two smaller triangles would be similar, yielding

   $$\frac{AC}{DC}=\frac{DC}{BC}.$$

   Plugging in $AC=a$ and $BC=b$ gives that $DC=\sqrt{ab}$ as desired.

   If you are interested in seeing the rules for compass and straightedge constructions, click on the link provided.