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	2013-03-22 17:14:55
Last modified on	2013-03-22 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 51-00
Related topic	ConstructionOfCentralProportion