Linear Algebra: Orthogonal Theorem presented in two different ways. Why and how is it possible? Book says:
According to the Pythagorean Theorem, two vectors are perpendicular if and only if:

Then, later is says: Let $u$ and $v$ be vectors in $\Bbb R^n$. Then u and v are orthogonal if and only if:

What I don't understand is how $\|v-u\|^2$ is the same as $\|u + v\|^2$.
Why does the book present orthogonal in two different yet similar formulas?
 A: In the image below, the first equation asks for the blue triangle to be right. The second equation asks for the pink triangle to be right. Since u and v are on the sides of both these triangles, this is all equivalent to $u \bot v$.

A: Just look at the dot product version:
We call two vectors $u$ and $v$ orthogonal if
$u_1v_1+u_2v_2 = 0$.
This, however is true precisely if $(-u_1)v_1+(-u_2)v_2=0$, that is, if $-u$ and $v$ are orthogonal. 
In other words, while in general $\|u-v\|^2\neq\|u+v\|^2$ it turns out that $\|u-v\|^2 = \|u\|^2+\|v\|^2$ precisely when $\|u+v\|^2=\|u-(-v)\|^2=\|u\|^2+\|-v\|^2=\|u\|^2+\|v\|^2$.
Actually, as a side remark, $\|u+v\|^2=\|u-v\|^2$ could serve as an alternative definition of orthogonality.
A: They are justifying the definition of orthogonality as $u\cdot v=0$.
With that definition, and the knowledge that $\|x\|^2=x\cdot x$, that computation can be rewritten:
$$
\|x-y\|^2=(x-y)\cdot(x-y)=x\cdot x-2(x\cdot y)+y\cdot y=\|x\|^2 -2(0)+\|y\|^2=\|x\|^2 +|y\|^2
$$
If you performed a similar computation with $\|x+y\|^2$:
$$
\|x+y\|^2=(x+y)\cdot(x+y)=x\cdot x+2(x\cdot y)+y\cdot y=\|x\|^2 +2(0)+\|y\|^2=\|x\|^2 +|y\|^2
$$
Draw the pictures of the vectors $x,y,x-y,x+y$ and slide them around to form the right trangles we are talking about.
