Prove that if the sum of $\vec{a}$ and $\vec{b}$ and the difference of vectors $\vec{a}$ and $\vec{b}$ are perpendicular Here's the full problem:
Prove that if the sum of $\vec{a}$ and $\vec{b}$ and the difference of vectors $\vec{a}$ and $\vec{b}$ are perpendicular, then the magnitude of $\vec{a}$ and $\vec{b}$ must be equivalent.
Okay, so i have tried going through the branch of the formula for perpendicular vectors sum $\ \vec{a}+\vec{b}=\sqrt{|\vec{a}|^2+|\vec{b}|^2}$$ but i don't think this works though.
I would like to see an approach without using dot prodcut of 2 perpendicular vectors being 0.
 A: 
You can use geometry. Vectors $\vec a$ and $\vec b$ form a paralelogram (as the Figure) and $\vec a+ \vec b$ and $\vec a- \vec b$ are the diagonals.
We know that these diagonals cut itself in the middle point. So $BI=DI$. If $CI$ is perpendicular to $BD$ then the triangles $BIC$ and $CID$ are congruents and then $BC=CD$.
A: You have, for the dot product $\langle .,. \rangle$,
$$\langle a+b, a-b \rangle = 0$$
so
$$\langle a, a\rangle -\langle a, b\rangle +\langle b, a\rangle -\langle b, b\rangle =0$$
and because $\langle a, b\rangle =\langle b ,a \rangle $, you deduce that
$$\langle a,a \rangle =\langle b , b\rangle $$
Edit : If you want another way of proving that, apply Pythagorean theorem to $a+b$ and $a-b$, which are perpendicular. You get that
$$|a+b+a-b|^2 = |a+b|^2+|a-b|^2$$
so
$$4|a|^2 = |a|^2+|b|^2+2 \langle a,b \rangle + |a|^2+|b|^2-2 \langle a,b \rangle$$
so $$|a|^2 = |b|^2$$
A: The dot product of sum and difference vectors vanish due to their being mutually  perpendicular.
$$(\vec a+\vec b)\cdot(\vec a-\vec b)=0$$
Multiply term by term
$$(\vec a\cdot\vec a)-(\vec b \cdot \vec b)-(\vec a\cdot\vec b)+(\vec a \cdot\vec b)=0$$
$$ |a|^2-|b|^2 =0$$
$$ |a|=|b|$$
EDIT1
Ok then geometrically with trig:
Slopes of sum, diff of vectors in a parallelogram
$$\tan \theta_1=\dfrac{b \sin \theta}{a+ b \cos \theta} $$
$$\tan \theta_2=\dfrac{b \sin \theta}{a- b \cos \theta} $$

Product of LHS tans is $-1$
$$\dfrac{b^2 \sin^2 \theta}{a^2- b^2 \cos^2  \theta}=-1$$
Cross multiply, transpose to simplify discard $a+b=0$ solution getting
$$a=b.$$
