A ratio that is golden

What do you see here? A triangle, for sure. With two circles, one with center E, the other with center C. The sides of the triangle are given in red, having lengths of 1, 2 and \sqrt{5}, according to the Pythagorean theorem a^2+b^2=c^2. As AE is the radius of the circle, DE is this radius as well, so DE is 1 (in green). It means that DC comprises of the rest, \sqrt{5}-1 (green as well).

A look at the other circle shows us that both BC and DC are its radius. As DC has length \sqrt{5}-1 which we just concluded, it goes for BC (green, below) as well. And as the length of AC is 2, AB must be the difference between AC and BC, so AB is equal to 2-(\sqrt{5}-1)=3-\sqrt{5}.

Let’s end this up with a bold claim: the ratio between AB and BC is the same as the ratio between BC and AB, or

    \[\frac{AB}{BC}=\frac{BC}{AC}\]

Is that so? Well, we can subsitute:

    \[\frac{3-\sqrt{5}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{2}\]

It means that

    \[(3-\sqrt{5})\cdot 2=(\sqrt{5}-1)\cdot (\sqrt{5}-1)\]

(cross multiplication)

which is

    \[2(3-\sqrt{5})=5-2\sqrt{5}+1\]

and that brings us to

    \[6-\sqrt{5}=6-\sqrt{5}\]

It seems we were right.

This  example is the most simple way to construct the Golden Ratio \frac{AC}{BC} (\approx 1.61803398875) with just a pair of compasses and a ruler.