Prove about a right triangle How to prove (using vector methods) that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices.
Defining the right triangle as the one formed by $\vec{v}$ and $\vec{w}$ with hypotenuse $\vec{v} - \vec{w}$, This imply to prove that $||\frac{1}{2}(\vec{v}+\vec{w})|| = ||\frac{1}{2}(\vec{v}-\vec{w})||$. 
The only thing that came to my mind is to expand like this:
\begin{align}
& {}\qquad \left\|\frac{1}{2}(\vec{v}+\vec{w})\right\| = \left\|\frac{1}{2}(\vec{v}-\vec{w})\right\| \\[10pt]
& =\left|\frac{1}{2}\right|\left\|(\vec{v}+\vec{w})\right\| = \left|\frac{1}{2}\right|\left\|(\vec{v}-\vec{w})\right\| \\[10pt]
& =\left\|(\vec{v}+\vec{w})\right\| = \left\|(\vec{v}-\vec{w})\right\| \\[10pt]
& =\left\|(v_1+w_1, v_2+w_2,\ldots,v_n+w_n)\right\| = \left\|(v_1-w_1, v_2-w_2,\ldots,v_n-w_n)\right\| \\[10pt]
& =\sqrt{(v_1+w_1)^2+ (v_2+w_2)^2+\cdots+(v_n+w_n)^2} \\[10pt]
& = \sqrt{(v_1-w_1)^2+ (v_2-w_2)^2+\cdots+(v_n-w_n)^2}
\end{align}
$$
And now I get stuck. Why? Because expanding it more would result in
$$v_1^2+2v_1w_1+w_1^2 \cdots = v_1^2-2v_1w_1+w_1^2 \cdots$$
And theres obiously a minus there.
I know a making a mistake but I can't get further. Any help?

If u and v are orthogonal, $u.v=0$.
Then $||u+v||^2 = ||u-v||^2 = ||u||^2+||v||^2$
