Unable to understand visual proof of Sophie Germain's Identity I've found a visual proof of Sophie Germain's Identity, however, I am unable to understand how the side lengths have been derived.
Visual proof of Sophie Germain's Identity
I understand the proof once the side lengths are found (just basic Pythagoras), however, I am unable to understand how they got the side lengths in the first place.
 A: You start with your rectangle $ABCD$ of dimensions $|b|^2 + |a - b||a + b|$ and $|b||a - b|$ (in the picture, we'll label clockwise from the top-left, so that $AB$ is the top edge, length $|b|^2 + |a - b||a + b|$). Why? You just do; it works in the end.
Next, construct the point $E$ on the bottom edge $CD$, so that $CE$ has length $|b|^2$ and $ED$ has length $|a - b||a + b|$. You similarly construct a point $F$ on the left edge $DA$, so that $DF$ has length $|b||a + b|$ and $FA$ has length $|b|||a + b| - |a - b||$. Note that, in order for this construction to work, we need $|a - b| \ge |a + b|$. I'm sure something similar could be achieved if $|a + b| \le |a - b|$, or indeed considering the same problem with $a' = a$ and $b' = -b$. Anyway, we assume $|a + b| \le |a - b|$ here for this picture.
Next, we note that $\triangle FDE$ and $\triangle ECB$ are similar triangles. They each share a right-angle, and further, their two legs are in proportion. We have
$$\frac{|EC|}{|CB|} = \frac{|b|^2}{|b||a - b|} = \frac{|b|}{|a - b|}.$$
Also,
$$\frac{|FD|}{|DE|} = \frac{|b||a + b|}{|a - b||a + b|} = \frac{|b|}{|a - b|}.$$
This shows that the two triangles are similar right-angled triangles. Thus, $\angle DFE = \angle CEB$ (both denoted $\beta$ in the diagram) and $\angle DEF = \angle CBE$ (denoted $\alpha$ in the diagram) add to $\pi/2$. We then have
$$\angle DEF + \angle FEB + \angle BEC = \angle DEC = \pi,$$
and $\angle DEF + \angle BEC = \pi/2$, so $\angle FEB = \pi/2$. This means that, if we draw a line from $F$ to $B$, the triangle $\triangle FEB$ is right-angled, and it's OK to use Pythagoras's theorem.
The rest, as you say, is just Pythagoras's theorem. We can determine the lengths of $FE$ and $EB$ using Pythagoras's theorem on, respectively $\triangle FDE$ and $\triangle ECB$. These hypotenuses are legs of the triangle $\triangle FEB$, which gives us the value of $|FB| = \sqrt{a^2 + 2ab + 2b^2}\sqrt{a^2 - 2ab + 2b^2}$. On the other hand, $\triangle FAB$ is also right-angled, so a separate application of Pythagoras's theorem shows that $|FB| = \sqrt{a^4 + 4b^4}$.
And so ends this interesting but rather perversely overly complicated proof of a very straightforward algebraic fact!
