Verify Archimedes’ construction for angle trisection

Verify Archimedes’ construction for trisecting an angle.

We are allowed to construct points using a straightedge and compass. Moreover, our straightedge has an (arbitrary) unit distance marked on it. We are permitted to ‘slide’ the marked straightedge on the plane.

Suppose we have an angle $\angle zxy$ of measure $\theta$, where $0 < \theta < \pi/2$. Using our unit ruler, we can say (without loss of generality) that $\overline{xy}$ has length 1. Construct the circle with center $x$ and containing $y$; this circle intersects the opposite ray of our angle at a point with we call (again without loss of generality) $z$. So $xz = 1$. Now extend the segment $\overline{xy}$ to a line.

Now slide the straightedge so that one mark is on the line $\overline{xy}$ and the other is on the circle centered at $x$. Finally, slide the straightedge (keeping the marks on the line and circle) so that the edge intersects point $z$. Construct this segment, calling the points where the edge intersects the line and circle $w$ and $u$, respectively, as in the following diagram.

In addition, label the angles $\alpha$, $\beta$, and $\gamma$ as shown.

Note that $\triangle xzu$ is isocoles, so that $\beta = \gamma$. Moreover, $\triangle uwx$ is isocoles. Now $\angle uxz$ has measure $\pi - \theta - \alpha = \pi - \gamma - \beta$, whence we have $\theta = 2\beta - \alpha$. Now $\angle wux$ has measure $\pi - 2\alpha$, and moreover $\beta + \pi - 2\alpha = \pi$, so that $\beta = 2\alpha$. Thus $\theta = 3\alpha$.