Geometric proof for $|| u ||^2 + || v ||^2 = \frac{1}{2}||u-v||^2 + 2||\frac{u+v}{2}||^2$

Question

Is there an geometric proof for the following identity?

$|| u ||^2 + || v ||^2 = \frac{1}{2}||u-v||^2 + 2||\frac{u+v}{2}||^2$.

The norm here is normal Euclidean norm, and $u,v$ are vectors.

Answer 1

Not sure what you mean by a "geometric proof" so I'll reduce it to a statement in classical Euclidean geometry.

Consider the vectors $u-v$ and $u+v$ to be two edges of a triangle which meet at the origin (which is a vertex of said triangle). Call this triangle $ABC$ where $A$ is the origin and $AB$ is $u-v$ and $AC$ is $u+v$.

In this case the vector $u$ is simply the median of this triangle. Call $D$ the midpoint of $BC$. The distances $BD$ and $CD$ are both half of $BC$ and are equal to $|| v||$.

So your formula turns into $\displaystyle \frac{AB^2 + AC^2}{2} = AD^2 + BD^2$ which is the well-known theorem of Apollonius.