Parallelogram law

Euclidean geometry

In Euclidean geometry, the parallelogram law states that for any parallelogram:

$$\sum \text{diagonal}^2 = \sum \text{side}^2$$

For example, in parallelogram $ABCD$:

$$AC^2+BD^2=AB^2+BC^2+CD^2+DA^2$$

Since $AB=CD$ $DA=BC$.

$$AC^2+BD^2=2(AB^2+BC^2)=2(CD^2+DA^2)$$

Generalisation into vector spaces

Let $V$ be an inner product space over $\mathbb{F}$.

$$\forall u,v\in V,\; \left\| u+v \right\|^2+\left\| u-v \right\|^2=2\left(\left\| u \right\|^2+\left\| v \right\|^2\right).$$

To illustrate this in $\mathbb{R}^2$:

Proof

By direct calculation, $$\begin{aligned} \| u+v \|^2 + \| u-v \|^2 &= \langle u+v, u+v \rangle + \langle u-v, u-v \rangle\\ &= \| u \|^2 + \| v \|^2 + \langle u, v \rangle + \langle v, u \rangle + \| u \|^2 + \| v \|^2 - \langle u, v \rangle -\langle v, u \rangle\\ &=2(\| u \|^2 + \| v \|^2). \end{aligned}$$