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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.

Illustration of u + v in R2

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

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    Is $u^2$ shorthand for $u \cdot u$? Then yes.2012-01-21

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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.

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    This criteria you can prove by considering areas of triangles ABO and ADO, where O is the midpoint of BD.2012-01-21