Geometrical progression from 1 to $\sqrt5$ to 3 by arithmetics

Question

How do you formulate arithmetically this pretty simple Euclid geometric progression from one to three via square root of 5?

GK = 1, AE = $\sqrt5$, GH = 3 but what is the equation behind the last line?

Answer 1

The two triangles $AIH$ ( $I$ is the intersection of $GH$ and $AE$) and $AKE$ are similar and so

$${IH\over AI}={AE\over AK}$$

We have $AE=\sqrt{5}$, $AK=GK=1$ and finally $AI={AE\over 2}={\sqrt{5}\over 2}$ and $GI={AK\over 2}={1\over 2}$. We deduce $IH={5\over 2}$ and so

$$\begin{align} GH&=IH+GI\\&={5\over 2}+{1\over 2}\\&=3\end{align}$$