Show that $u+v$ bisects $u$ and $v$ only if $|u|=|v|$ I want to show that if I have two Euclidian vectors in $\mathbb{R}^n$ than the sum of these two vectors bisects the angle between the two vectors. Said more mathematically. 
Let  $(u,v) \in \mathbb{R}^n$
Then $\angle(u,v+u) = \angle(u+v,v)$ if and only if $|u|=|v|$
I tried using the fact that 
$$ \angle(u,v) = \arccos \left( \frac{u \cdot v}{|u| |v|} \right) $$
Alas this attempt was futile
Now for $\mathbb{R}^2$ this is obviously true, as can bee seen from my illustration. 

How do I prove this with some rigor? Thanks for all tips and advices, this is not a homework question.  

The first prove:
Two vectors form a parallelogram ABCD, we know that in a parallelogram diagonals bisect by their point of intersection. It means that segment from A to point of intersections of diagonals is a middle line of triangle ABD. There is a criteria that middle line is bisecting line if and only if the triangle is isosceles. That proves your problem.
The second prove:
Let $|u|=|v|$ is equal $u^2=v^2$ then $\cos(\phi_1)=\frac{(u+v,u)}{|u+v||u|}=\frac{u^2+(u,v)}{|u+v||u|}=\frac{v^2+(u,v)}{|u+v||v|}=\cos(\phi_2)$
Let $\cos(\phi_1)=\cos(\phi_2)$ then $\frac{u^2+(u,v)}{|u|}=\frac{v^2+(u,v)}{|v|}$ => $|u|+\frac{(u,v)}{|u|}=|v|+\frac{(u,v)}{|v|}$=>$(|u|-|v|)(1-\frac{(u,v)}{|u||v|})=0$ if $u$ and $v$ aren't collinear, it means that $|u|=|v|$ But if they are, it means that angle is zero. And angle $\phi_1$ and $\phi_2$ are also zero.
