Vector proof that $d_1^2 + d_2^2 = 2a^2 + 2b^2$ in a parallelogram

Question

How would one prove the equality of the sum of squares of diagonals and twice the sum of squares of the two sides:

$$\left|\mathbf{p} + \mathbf{q}\right|^2 + \left|\mathbf{p} - \mathbf{q}\right|^2 = 2\left|\mathbf{p}\right|^2 + 2\left|\mathbf{q}\right|^2 $$

where $\mathbf{p}$ and $\mathbf{q}$ are vectors, representing two intersecting sides of a parallelogram.

Am I supposed to take the LHS and prove it equals the RHS like in a normal proof?

Answer 1

The key is to use the dot product to simplify the proof. Remember that $\left|\vec{v}\right|^2=\vec{v} \cdot \vec{v}$.

$$ \begin{align} \left| \vec{p} + \vec{q} \right|^2 + \left|\vec{p} - \vec{q} \right|^2 &= \left(\vec{p} + \vec{q}\right) \cdot \left(\vec{p} + \vec{q}\right) + \left(\vec{p} - \vec{q}\right) \cdot \left(\vec{p} - \vec{q}\right) \\ &= \left(\vec{p} \cdot \vec{p} + 2 \vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{q}\right) + \left(\vec{p} \cdot \vec{p} - 2 \vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{q}\right) \\ &= 2 \vec{p} \cdot \vec{p} + 2 \vec{q} \cdot \vec{q} \\ &= 2 \left|\vec{p}\right|^2 + \left|\vec{p}\right|^2 \end{align} $$

Answer 2

$$|p+q|^{2} = \langle p+q, p+q\rangle = \langle p, p+q\rangle + \langle q, p+q\rangle = \langle p,p\rangle + \langle p,q\rangle + \langle q,p\rangle + \langle q, q\rangle $$

Repeat for $|p-q|^{2}$ and add.