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Hometrisection of angle
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trisection of angle
Given an angle of measure $\alpha$ such that $0<\alpha\leq\frac{\pi}{2}$, one can construct an angle of measure $\frac{\alpha}{3}$ using a compass and a ruler with one mark on it as follows:
1. Construct a circle $c$ with the vertex $O$ 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 $\angle CDO$ 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 $\angle AOB$ is an exterior angle of $\triangle BOD$, we have that $m(\angle AOB)=m(\angle OBD)+m(\angle ODB)$;

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

Since $\angle OCB$ is an exterior angle of $\triangle OCD$, we have that $m(\angle OCB)=m(\angle COD)+m(\angle CDO)$;

Note that $\angle OBC=\angle OBD$ and $\angle ODB=\angle CDO$;

Thus,
$\begin{array}[]{rl}\alpha&=m(\angle AOB)\\ &=m(\angle OBD)+m(\angle ODB)\\ &=m(\angle OBC)+m(\angle CDO)\\ &=m(\angle OCB)+m(\angle CDO)\\ &=m(\angle COD)+m(\angle CDO)+m(\angle CDO)\\ &=3m(\angle CDO).\end{array}$
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 $0<\beta\leq 2\pi$:

If $0<\beta\leq\frac{\pi}{2}$, then use the construction given above.

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

If $\pi<\beta\leq\frac{3\pi}{2}$, then trisect an angle of measure $\beta\pi$ and add on an angle of measure $\frac{\pi}{3}$ to the result.

If $\frac{3\pi}{2}<\beta\leq 2\pi$, 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.
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