How can we get Pythagoras from the Parallelogram law?

Question

When using the definition and properties of the inner product, we get the parallelogram law: $$\:\,\|x+y\|^2 =\: \langle x+y, x+y\rangle= \langle x, x\rangle + \langle x, y\rangle +\langle y, x\rangle +\langle y, y\rangle\,,\\ \|x-y\|^2 =\:\langle x-y, x-y\rangle= \langle x, x\rangle - \langle x, y\rangle -\langle y, x\rangle +\langle y, y\rangle $$

Adding these two expressions: $$\|x+y\|^2+\|x-y\|^2 \:=\: 2\langle x, x\rangle + 2\langle y, y\rangle \:=\: 2\|x\|^2+2\|y\|^2$$ as required. But the above does not simplify to Pythagoras' theorem $\|x+y\|^2=\|x\|^2+\|y\|^2$
if $x$ and $y$ are orthogonal.

How can we get Pythagoras from the parallelogram law?

Answer 1

If you are talking strictly about the parallelogram, when the vector are orthogonal (hence $\langle x, y\rangle=0$, you have

$||x+y||^2= \langle x+y, x+y\rangle= \langle x, x\rangle + \langle x, y\rangle +\langle y, x\rangle +\langle y, y\rangle=\|x\|^2+\|y\|^2$

You can also see it "geometrically", by noticing that this means that when $x$ and $y$ are orthogonal, you have a rectangle, with both diagonals of same length, that is

$||x+y||^2=||x-y||^2$...