The following question is one of the exercises "Foundation Euclidean and non-Euclidean geometry" by Greenberg (chapter 1/ Major Exercises/ 3 )
For any angle, draw a circle $\gamma$ radius $d$ centered at the vertex $O$ of the angle. This circle cuts the sides of the angle at points $A$ and $B$. Place the marked straightedge so that one mark gives a point $C$ on line $\overline{OA}$ such that $O$ is between $C$ and $A$, the other mark gives a point $D$ on circle $\gamma$, and the straightedge must simultaneously rest on the point $B$, so that $B$, $C$, and $D$ are collinear (the following Figure). Prove that $\angle COD$ so constructed is one-third
of $\angle AOB$. (Hint: Use Euclidean theorems on exterior angles and isosceles
triangles.)
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Thanks for any help.
Let $\angle COD = \theta$
As $\triangle COD$ is isosceles, $\angle DCO = \theta$ and $\angle CDO = \pi - 2\theta$.
$\triangle DOB$ is isoceles, with base angles at vertices $D$ and $B$. Hence apex $\angle DOB = \pi - (2)(\pi - 2\theta) = 4\theta - \pi$
$\angle AOD = \pi - \theta$
Hence $\angle AOB = \pi - \theta + 4\theta - \pi = 3\theta$
$\therefore \angle COD = \frac 13 \angle AOB \ (QED)$