## Problem on proving the parallelogram law with vectors

Prove the parallelogram law: The sum of the squares of the lengths of both diagonals of a parallelogram equals the sum of the squares of the lengths of all four sides.
• ## Solution

#### Posing the parallelogram law precisely

Let's locate a corner of the parallelogram at the origin. We let the neighboring two vertices be given by the vectors $\bfa$ and $\bfb$.
We now express the diagonals in terms of $\bfa$ and $\bfb$.
Recall that
The diagonals are given by $\bfa + \bfb$ and $\bfb - \bfa$:
We can now formulate the parallelogram law precisely:
The sum of the squares of the lengths of the diagonals is $$\left| \bfa + \bfb \right|^2 + \left| \bfb - \bfa \right|^2.$$
The sum of the squares of the lengths of the sides is $$2 \left| \bfa \right|^2 + 2 \left| \bfb \right|^2.$$
Hence, we are to show that $$\left| \bfa + \bfb \right|^2 + \left| \bfb - \bfa \right|^2 = 2 \left| \bfa \right|^2 + 2 \left| \bfb \right|^2.$$

#### Proving the parallelogram law

Recall that
We can compute the value of the left hand side:\begin{align}
\left| \bfa + \bfb \right|^2 + \left| \bfb - \bfa \right|^2 &= (\bfa + \bfb) \cdot (\bfa + \bfb) + (\bfb - \bfa) \cdot (\bfb - \bfa)
\end{align}
Cancelling the $\bfa\cdot\bfb$ terms and using the relationship of dot product to vector length again, we get \begin{align}
\left| \bfa + \bfb \right|^2 + \left| \bfb - \bfa \right|^2 &= 2 \ \bfa \cdot \bfa + 2 \ \bfb\cdot \bfb\\
&= 2 \left|\bfa\right|^2 + 2 \left| \bfb \right|^2,
\end{align} which is what we sought to prove.