Is there a geometric interpretation to the Brahmagupta–Fibonacci identity?

Question

The Brahmagupta–Fibonacci identity says that, for integers $a,b,p,q$:

$(a^2+b^2)(p^2+q^2)=(ap+bq)^2+(aq-bp)^2$

Is there an intuitive geometric way to interpret this?

Thanks!

Answer 1

@Mindlack has given a straightforward algebraic proof.

Here is another proof, geometricaly oriented (indeed, it involves both the dot product and the cross product):

Let $u=\binom{a}{b}$ and $v=\binom{p}{q}$.

$$\|u\|^2\|v\|^2=(u.v)^2+(\|u \times v\|)^2$$

which is true because it boils down to:

$$\|u\|^2\|v\|^2=(\|u\|\|v\|\cos \theta)^2+((\|u\|\|v\|\sin \theta)^2$$

(a consequence of $(\cos \theta)^2+(\sin \theta)^2=1$)

Answer 2

I assume this one counts as geometric.