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.
Title  compass and straightedge construction of geometric mean 

Canonical name  CompassAndStraightedgeConstructionOfGeometricMean 
Date of creation  20130322 17:14:55 
Last modified on  20130322 17:14:55 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  10 
Author  Wkbj79 (1863) 
Entry type  Algorithm 
Classification  msc 51M15 
Classification  msc 5100 
Related topic  ConstructionOfCentralProportion 