compass and straightedge construction of geometric mean

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 $C$.

2. Extend the line segment past $C$.

3. Mark off a line segment of length $b$ such that one of its endpoints is $C$. Label its other endpoint as $B$.

4. Construct the perpendicular bisector of $\overline{AB}$ in order to find its midpoint $M$.

5. Construct a semicircle with center $M$ and radii $\overline{AM}$ and $\overline{BM}$.

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.

   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 provided.