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?

Suppose the vertex opposite the hypotenuse is at the origin. If we choose another vertex v then we can think of the vector from the midpoint 1/2(v + w) to v as one half of the hypotenuse. 
If we transport this vector from its base at 1/2( v + w) to the origin, the base of this new vector is at the origin and its other point is at v-1/2(v + w). v-1/2(v + w) is 1/2(v - w). Now this new vector is like the vector 1/2(v + w) with the exception that it except that the instead of adding 1/2 w to 1/2 w we are adding -1/2 w. 
This means that if we choose 1/2 v as the direction of an axis then the vector we just transported will fall the same amount to the perpendicular axis as the vector from our origin to the midpoint. This means they are same size.
