The diagonal … has

The diagonal $PR$ has $|PR{|}^2=(\mathbf{A}+\mathbf{B})\cdot (\mathbf{A}+\mathbf{B})=\mathbf{A}\cdot \mathbf{A}+\mathbf{A}\cdot \mathbf{B}+\mathbf{B}\cdot \mathbf{A}+\mathbf{B}\cdot \mathbf{B}$. Add $|QS{|}^2$ from the other diagonal to prove the parallelogram law: $|PR{|}^2+|QS{|}^2=$ sum of squares of the four side lengths.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §11.1 Vectors and Dot Products, #32