Ytukyg

How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.

https://www.youtube.com/watch?v=9VceAA2qBPA

After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.

By the Pythagorean theorem,
$$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
$$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$

It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
$$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.

Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$