TITLE: Geometric proof Brahmagupta-Fibonacci identity
QUESTION [9 upvotes]: The Brahmagupta-Fibonacci identity states that for all $a,b,c,d$, $$(a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2.$$ 
This is easy to prove algebraically, and says essentially that the norm on complex numbers is multiplicative. Is there a geometric proof though? Bak and Newman's textbook "Complex Analysis" suggests there is (p.19), but I can't see how it would go.

REPLY [5 votes]: One geometric proof is shown here: