Proof of Construction of regular pentagon by using compass and straightedge.

Question

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

Constructing a Regular Pentagon (Video on YouTube)

Answer 1

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, $$|CD| = \sqrt{|OD|^2 + |OC|^2} = \sqrt{(\sqrt{5}-1)^2+2^2} = \sqrt{10-2\sqrt{5}}$$

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

Answer 2

Here is a proof that depends on the diagonal/side ratio in a regular pentagon equaling the golden ratio $\phi$, which probably would have been familiar to the Greeks who developed the construction (specifically Ptolemy -- which gives a subtle clue about one of its main premises).

The proof uses the picture below, modified from the answer given by Blue. I add the point on the circle diametrically opposite C, labeled C', and I label an additional vertex of the pentagon F. The slender blue lines define an inscribed triangle CED and an inscribed quadrilateral CEC'F. The objective is to prove that the triangle, isosceles by construction, has a base/leg ratio equal to $\phi$ making its vertices fit those of a regular pentagon and thus assuring that the apex angle ECF matches the required vertex angle.

We begin by defining the golden ratio in this construction.

Lemma: In a right triangle having legs $a$ and $b=2a$ and hypotenuse $c$, the ratio $b/(c-a)$ is golden.

We render the Pythagorean Theorem as

$b^2=c^2-a^2=(c+a)(c-a),$

where the factorization is proved by dissecting a rectagle of dimensions $c+a,c-a$ into a square minus a smaller square (this is how the Greeks understood it). Then with $b=2a$:

$b^2=(c+a)(c-a)=[(c-a)+b](c-a)$

and from the cross-product rule of proportions:

$\dfrac{(c-a)+b}{b}=\dfrac{b}{c-a},$

which corresponds to a golden sectioning of $c+a=(c-a)+b$. Lemma proved.

So if the radius of the circle is $r$, then OD measures $r/\phi$ and from this, CE measures $r\sqrt{1+(1/\phi)^2}$.

Triangle CC'E has a right angle (inscrived in a semicircle) at E, so the Pythagorean Theorem renders

$C'E=r\sqrt{4-(1+1/\phi^2)}$

and we use the fact (again known since ancient times) that powers of the golden ratio satisfy what we now call the Fibonacci relation $\phi^{n+2}=\phi^{n+1}+\phi^n$:

$C'E=r\sqrt{3-1+1/\phi}$

$=r\sqrt{2+\phi-1}$

$=r\sqrt{1+\phi}=r\phi.$

The same will apply to C'F by symmetry.

So now, by Ptolemy's Theorem (the product of diagonals of an inscribed convex quadrilateral equals the sum of products of the pairs of opposing sides):

$(EF)(2r)=(CE)(r\phi)+(CF)(r\phi)$

where by construction, CE and CF are identical, and so

$EF/CE=EF/CF=\phi,\text{ qed.}$