Pythagorean Triangle from Fibonacci Numbers

Theorem
Take $4$ consecutive Fibonacci numbers:
 * $F_n, F_{n + 1}, F_{n + 2}, F_{n + 3}$

Let:
 * $a := F_n F_{n + 3}$
 * $b := 2 F_{n + 1} F_{n + 2}$
 * $c := F_{2 n + 3}$

Then:
 * $a^2 + b^2 = c^2$

and:
 * $\dfrac {a b} 2 = F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}$

That is, if the legs of a right triangle are the product of the outer terms and twice the inner terms, then:
 * the hypotenuse is the Fibonacci number whose index is half the sum of the indices of the four given Fibonacci numbers.
 * the area is the product of the four given Fibonacci numbers.

Proof
By definition of Fibonacci numbers:
 * $F_n = F_{n + 2} - F_{n + 1}$

and:
 * $F_{n + 3} = F_{n + 2} + F_{n + 1}$

Then $a$ can be expressed as:

Now consider:

and it can be seen that the ordered triple:
 * $\paren {{F_{n + 2} }^2 - {F_{n + 1} }^2, 2 F_{n + 1} F_{n + 2}, {F_{n + 2} }^2 + {F_{n + 1} }^2}$

is in the form of Solutions of Pythagorean Equation.

It remains to be shown that:
 * ${F_{n + 2} }^2 + {F_{n + 1} }^2 = F_{2 n + 3}$

We have:

and the proof is complete.

The area is trivial: