# 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. 1.

Draw a line segment of length $a$. Label its endpoints $A$ and $C$.

2. 2.

Extend the line segment past $C$.

3. 3.

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

4. 4.

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

5. 5.

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

6. 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 CompassAndStraightedgeConstructionOfGeometricMean 2013-03-22 17:14:55 2013-03-22 17:14:55 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Algorithm msc 51M15 msc 51-00 ConstructionOfCentralProportion