trisection of angle
Given an angle of measure (http://planetmath.org/AngleMeasure) $\alpha $ such that $$, one can construct an angle of measure $\frac{\alpha}{3}$ using a compass and a ruler (http://planetmath.org/MarkedRuler) with one mark on it as follows:

1.
Construct a circle $c$ with the vertex (http://planetmath.org/Vertex5) $O$ of the angle as its center. Label the intersections^{} of this circle with the rays of the angle as $A$ and $B$. Mark the length $OB$ on the ruler.

2.
Draw the ray $\overrightarrow{AO}$.

3.
Use the marked ruler to determine $C\in c$ and $D\in \overrightarrow{AO}$ such that $CD=OB$ and $B$, $C$, and $D$ are collinear^{}. Draw the line segment^{} $\overline{BD}$. Then the angle measure of $\mathrm{\angle}CDO$ is $\frac{\alpha}{3}$. (The line segment $\overline{OC}$ is drawn in red. Having this line segment drawn is useful for reference purposes for the justification of the construction.)
Let $m$ denote the measure of an angle. Then this construction is justified by the following:

•
Since $\mathrm{\angle}AOB$ is an exterior angle^{} of $\mathrm{\u25b3}BOD$, we have that $m(\mathrm{\angle}AOB)=m(\mathrm{\angle}OBD)+m(\mathrm{\angle}ODB)$;

•
Since $OC=OB=CD$, we have that $\mathrm{\u25b3}BOC$ and $\mathrm{\u25b3}OCD$ are isosceles triangles^{};

•
Since the angles of an isosceles triangle are congruent^{}, $m(\mathrm{\angle}OBC)=m(\mathrm{\angle}OCB)$ and $m(\mathrm{\angle}COD)=m(\mathrm{\angle}CDO)$;

•
Since $\mathrm{\angle}OCB$ is an exterior angle of $\mathrm{\u25b3}OCD$, we have that $m(\mathrm{\angle}OCB)=m(\mathrm{\angle}COD)+m(\mathrm{\angle}CDO)$;

•
Note that $\mathrm{\angle}OBC=\mathrm{\angle}OBD$ and $\mathrm{\angle}ODB=\mathrm{\angle}CDO$;

•
Thus,
$\begin{array}{cc}\hfill \alpha & =m(\mathrm{\angle}AOB)\hfill \\ & =m(\mathrm{\angle}OBD)+m(\mathrm{\angle}ODB)\hfill \\ & =m(\mathrm{\angle}OBC)+m(\mathrm{\angle}CDO)\hfill \\ & =m(\mathrm{\angle}OCB)+m(\mathrm{\angle}CDO)\hfill \\ & =m(\mathrm{\angle}COD)+m(\mathrm{\angle}CDO)+m(\mathrm{\angle}CDO)\hfill \\ & =3m(\mathrm{\angle}CDO).\hfill \end{array}$
Note that, since angles of measure $\frac{\pi}{6}$, $\frac{\pi}{3}$, and $\frac{\pi}{2}$ are constructible using compass and straightedge, this procedure can be extended to trisect any angle of measure $\beta $ such that $$:

•
If $$, then use the construction given above.

•
If $$, then trisect an angle of measure $\beta \frac{\pi}{2}$ and add on an angle of measure $\frac{\pi}{6}$ to the result.

•
If $$, then trisect an angle of measure $\beta \pi $ and add on an angle of measure $\frac{\pi}{3}$ to the result.

•
If $$, then trisect an angle of measure $\beta \frac{3\pi}{2}$ and add on an angle of measure $\frac{\pi}{2}$ to the result.
This construction is attributed to Archimedes.
References
 1 Rotman, Joseph J. A First Course in Abstract Algebra. Upper Saddle River, NJ: PrenticeHall, 1996.
Title  trisection of angle 

Canonical name  TrisectionOfAngle 
Date of creation  20130322 17:16:35 
Last modified on  20130322 17:16:35 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  11 
Author  Wkbj79 (1863) 
Entry type  Algorithm 
Classification  msc 01A20 
Classification  msc 51M15 
Related topic  VariantsOnCompassAndStraightedgeConstructions 