I have to prove the following statement:
If a regular $n$-gon is constructible, then so is a regular $2n$-gon.
My Attempt: 1. Draw a point at the each vertex. 2. Draw a line between each point. -> This will result in an intersection at the centre. Draw a point called $O$. And the line between the point and a vertex will be some distance $d$. -> Also, there will be $n$ angles $\theta = 2\pi/n$. So $\theta$ is constructible. 4. Now, we can construct a circle with radius $d$, at origin $O$. 3. If $\theta$ is constructible, so is $\phi = \theta/2$ (a result that is previously shown in assignment, no need for proof). 4. For each line from $O$ draw a line separated by $\phi$. -> This line will pass through $O$ and the circle. Draw points on each intersection of the line with the circle. 5. Draw a line each vertices and these newly drawn points.
Questions / Need substance: 1. Can I just assume that a point can be drawn at each vertex? 2. The centre point and the angles seem pretty obvious, but I'm not sure how I can justify this/if I need to.
At a higher level, you can apply the early result that a number is constructible iff it generates an extension of $\Bbb Q$ that has degree a power of $2$.
Constructing an $n$-gon amounts to being able to construct $\cos(\theta)$ where theta is the interior angle.
The half-angle identity says
$$\cos(\frac\theta 2)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}$$
This makes it clear that $\cos(\theta/2)$ lies in an extension of degree two or less of our power-of-2 extension, and that means the half angle is constructible too, and so is the 2n-gon.
As for the manual construction, I would say that you circumscribe the n-gon with a circle, bisect the interior angles (this is easy: you know it, right?) then treat the n new points where the bisecting meet the circle as the new vertices needed for the 2n-gon.