Show that a diagonal in a pentagon is the golden ratio

Question

I just learned that the diagonal of a pentagon (size 1) is the golden ratio (https://twitter.com/fermatslibrary/status/1210561047154872320)

I tried to verify that, but ended up having to show that: $$\cos{\frac{2\pi}{5}}=\frac{1}{\phi}$$

Question: is there a way to calculate the diagonal of a pentagon without having to do any relatively complex trigonometric calculations? For example only by drawing some smart supportive lines and using the Pythagorean Theorem?

Answer 1

Consider a regular pentagon $ABCDE$ and a vertex inscribed pentagram $ACEBD$ with the additional intersection points $a,...,e$ somewhat closer to the center but on the oposite ray. Then the pentagram will have sides $AdeC, CabE, EcdB, BeaD, DbcA$.

By assumption you have $AB=BC=CD=DE=EA=1$. Let further be $Ad=Bd=Be=Ce=Ca=Da=Db=Eb=Ec=Ac=:x$ and $ab=bc=cd=de=ea=:y$.

Now consider the isoscele triangle $AEd$. Thus you get $1=x+y$. Its base is $x$. Then consider the scaled down isoscele triangle $Ebc$ with sides $x, x, y$ (its tip angle clearly is the same). From those you get the scaling ratio $$\varphi:=\frac1x=\frac xy$$ Together with the above ($y=1-x$) the equation for the golden ratio follows: $$1-x=x^2$$ or, when dividing by $x^2$ and re-inserting $\varphi$: $$\varphi^2=\varphi+1$$ --- rk

Answer 2

$$\triangle AA'C\sim\triangle BB'C\quad\to\quad\frac{|A'C|}{|A'A|}=\frac{|B'C|}{|B'B|}\quad\to\quad \underbrace{\frac{a+b}{a}=\frac{a}{b}}_{=\,\phi\;\text{(by def'n)}} \quad\to\quad \frac{\text{diagonal}}{\text{side}}= \phi$$