Golden Ratio Conjecture in three simple Geogebra shapes--circle, triangle, and square. A circle, equilateral triangle, and square of equal heights are all placed on the same horizontal line as shown below.  The circle is tangent to the triangle which is centered upon the left edge of the square.  A line is drawn from the center of the circle to the right edge of the square, passing through the center of the square.  The line is cut into two segments by the right side of the triangle, as shown.
Show that the ratio of the length of the blue segment to the green segment is the golden ratio 1.618.  (Is it? It seems so!) 

I have been playing around in geogebra, but I was unable to get the circle tangent to the triangle as shown in the figure, which I drew in Adobe Illustrator.  Any geogebra assistance would be appreciated!  How do I move/translate a simple object in such a manner?  I am used to dragging and dropping it in Adobe Illustrator, but geogebra is much better suited to these golden ratio conjectures.  Thanks for all your help!  I have been successful with geogebra with a couple other constructions which I will share soon. :)  

Here is a diagram of the situation.

Let us calculate the relevant lengths, assuming that the radius of the circle is $1$ so that we don't have to deal with scaling. One should note that $\angle CAB$ is $\pi/6$, equal to $\angle CDE$. Thus, the length of $AC$ is $\frac{1}{\cos(\pi/6)}$. The length of $CE$ and $EF$ separately is $\tan(\pi/6)$. Noting that $\cos(\pi/6)$ is $\sqrt{3}/2$ and $\tan(\pi/6)=\frac{1}{\sqrt{3}}$, we find that the length from $A$ to $F$ is $\frac{4}{\sqrt{3}}$. The length from $F$ to $G$ is $2-\frac{1}{\sqrt{3}}$. The ratio of these
$$\frac{\frac{4}{\sqrt{3}}}{2-\frac{1}{\sqrt{3}}}=\frac{4}{2\sqrt{3}-1}$$
which is not the golden ratio, but is $1.623$ instead.
