Proving $||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}|| \iff \vec{a} \perp \vec{b}$ Have some  non-null $\vec{a}$ and $\vec{b}$.


I am trying to prove this to no avail:
$$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}|| \iff \vec{a} \perp \vec{b}$$

If we start with
$$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}|| \implies \vec{a} \perp \vec{b}$$
Our hypothesis is
$$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}||$$
Which tells us that both horizontal sides of this triangle have the same length, so we got an isosceles triangle. Not sure what to make out of that though.
Anyway, the hypothesis is equivalent to
$$\sqrt{(\vec{a} + \vec{b})\cdot (\vec{a} + \vec{b})} = \sqrt{(\vec{a} - \vec{b})\cdot (\vec{a} - \vec{b})}$$
I can't make much out of that.

A hint to begin tackling this problem would be appreciated.
 A: Note that
$$
||\vec{a}+\vec{b}||^2 -||\vec{a}-\vec{b}||^2=4 \vec{a}\cdot \vec{b}
$$
and your conclusion follows.
A: Hint : take squares and use $||u+v||^2=||u||^2+||v||^2+2(u|v)$
A: If you square both sides we have:
$$\lvert\lvert a+b\rvert\rvert^{2}=\lvert\lvert a-b\rvert\rvert^{2}$$
so by expanding and removing equal terms we get
$$4a\cdot b=0$$
A: The way I would solve this would be by squaring (like the other answers).
However, if you want to proceed your geometric thought, you can see that the 2 small triangles have ALL sides equal:


*

*$||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}||$

*$||\vec{b}|| = ||-\vec{b}||$

*$||\vec{a}|| = ||\vec{a}||$


If 2 triangles have 3 sides equal, then they are equal.
Since the 2 triangles are equal, so are their angles. So, $\vec{a} \perp \vec{b}$
A: $||\vec{a} + \vec{b}||$, and $||\vec{a} - \vec{b}||$can be interpreted as the length of the longer and shorter diagonals of a parallelogram whose sides are $||\vec{a}||$, and $||\vec{b}||$. So if they are equal, the parallelogram becomes a rectangle,and so the angle between $\vec{a}$, and $\vec{b}$ must be a right angle. Thus: $\vec{a} \perp \vec{b}$.
