I'd like to know reasoning behind the diagram below. How is it possible, that this geometric progression gives right (I suppose) value for the equation:
$$\sqrt{2\sqrt2-1}$$
I understand $2\sqrt2 - 1$ from the diagonal of a square, but then logic further remains mystery to me.
- Radius of the circle IJ is $1$
- Diagonal IC is $2\sqrt2$
- JC is $2\sqrt2 -1$
Then I have to split radius of the circle JC and find intersecting point of the half of the circle JC from vertical line passing at point -1. That will give the final value for $\sqrt{2\sqrt2-1}$. But what is happened here, Id like to apply this further.
Everything just follows from Euclid's theorem stating that, if $ABC$ is a triangle with $\widehat{BAC}=\frac{\pi}{2}$ and $D$ is the projection of $A$ on $B$, $$ AD^2 = DB\cdot DC.$$ That is the reason for which, if we are able to construct some quantity $\alpha>0$ by straightedge and compass, we are able to construct $\sqrt{\alpha}$ as well. It is enough to take $DB=1$, $DC=\alpha$ and depict a semicircle $\Gamma$ over the diameter $BC$. If $A$ is the intersection between $\Gamma$ and the perpendicular to $BC$ through $D$, $AD=\sqrt{\alpha}$ as wanted.