I am trying to prove the converse of the Pythagorean theorem using vector geometry and vector algebra. Here's a picture:

Take notice that $\displaystyle \vec{c} = \vec{a} + \vec{b}$. The given is that

$\displaystyle |\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} + \vec{b}|^2$

The goal is to show that $\displaystyle v = 90^\circ$. I dotted $\displaystyle \vec{a}$ with $\displaystyle \vec{b}$, thus yielding

$\displaystyle \vec{a}\cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos v$

I then attemped to find an expression for $\displaystyle \vec{a}\cdot \vec{b}$ from the equation $\displaystyle \vec{a} + \vec{b} = \vec{c}$ by multiplying both sides by $\displaystyle \vec{a}$ and dot every vector with it.

$\displaystyle \vec{a}(\vec{a} + \vec{b}) = \vec{c} \cdot \vec{a}$

$\displaystyle |\vec{a}|^2 + |\vec{b}| \cdot |\vec{a}| \cdot \cos v = |\vec{c}|\cdot |\vec{a}| \cdot \cos \theta $

$\displaystyle \theta$ is the angle formed between $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$. I then I figured that I could compare the equation with $\displaystyle |\vec{a}|^2 + |\vec{b}|^2 = |\vec{a} + \vec{b}|^2$ and then identify the terms. This however did not work out as it lead to showing that $\displaystyle \vec{a}$ is the hypotenuse whereas $\displaystyle \vec{c}$ is not, thus contradicting the very premise of the proof.

Anyone got an idea?

PS: I've already seen the proof using congruence but that particular proof is rather irrelevant, so don't bother referencing it.