Prove A perpendicular to B if |a-b| = |a+b| I'm absolutely exhausted working on this problem. I think I am very close, but I don't know what to do from here. I've tried a lot of different algebraic approaches. 
So the problem says to prove A is perpendicular to B if |a-b| = |a+b|(we are talking about vectors and this is calculus class and this unit's recent lessons include the dot product and cross product).
What I've tried so far:
I've tried drawing a diagram and using simple vector math and the property of a right triangle I noticed that if A and B are set up as connecting vectors at a right angle, then |a+b| would be the hypotenuse and |a-b| would be the hypotenuse of a mirrored right angle triangle. From there I tried plugging it into the Pythagorean formula as such:
C^2 = A^2 + B^2    since this would prove A and B are perpendicular
Since C = |a+b| = |a-b| I substituted in...
|A+B||A-B| = |A|^2 + |B|^2
|A^2-B^2| = |A|^2 + |B|^2
I've been working on this for hours trying rearranging and I think I am need to try something new. I realize the dot product will prove perpendicular, but I don't know how to link the two. I'm stumped; 
Any help is appreciated.
 A: $$\|\vec{a} + \vec{b}\| = \|\vec{a} - \vec{b}\|$$
Dotting a vector with itself is the square of its length:
$$\|\vec{a} + \vec{b}\|^2 = \|\vec{a} - \vec{b}\|^2$$
$$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$$
Dot product distributes over addition:
$$\vec{a} \cdot (\vec{a} + \vec{b}) + \vec{b} \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot (\vec{a} - \vec{b}) - \vec{b} \cdot (\vec{a} - \vec{b})$$
$$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$$
$$2 (\vec{a} \cdot \vec{b}) = -2 (\vec{a} \cdot \vec{b})$$
$$\vec{a} \cdot \vec{b} = 0$$
Vectors dot to $0$ iff they are perpendicular: $\vec{a}$ is perpendicular to $\vec{b}$.
A: I assume you are working with real inner products. Since $\|a-b\|=\|a+b\|$, we have $\langle a-b,a-b\rangle=\langle a+b,a+b\rangle$. But 
$$\langle a-b,a-b\rangle = \langle a,a-b\rangle - \langle b,a-b\rangle =\langle a,a\rangle - \langle a,b\rangle - \langle b,a\rangle + \langle b,b\rangle$$
and
$$\langle a+b,a+b\rangle = \langle a,a+b\rangle + \langle b,a+b\rangle =\langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle + \langle b,b\rangle$$
so subtracting the two right-hand sides gives us $\langle a,b\rangle + \langle b,a\rangle=0$. What can you conclude from this?
A: Although the vector algebra in the other answers is probably what was intended when the problem was posed, I can't resist pointing out that, as you might have noticed when you drew pictures, the problem amounts to the theorem, from Euclidean geometry, that if the diagonals of a parallelogram have the same length then the parallelogram is a rectangle.  
