Parallelogram law

In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. We use these notations for the sides: AB, BC, CD, DA. But since in Euclidean geometry a parallelogram necessarily has opposite sides equal, that is, AB = CD and BC = DA, the law can be stated as $${\displaystyle 2AB^{2}+2BC^{2}=AC^{2}+BD^{2}\,}$$

If the parallelogram is a rectangle, the two diagonals are of equal lengths AC = BD, so $${\displaystyle 2AB^{2}+2BC^{2}=2AC^{2}}$$ and the statement reduces to the Pythagorean theorem. For the general quadrilateral (with four sides not necessarily equal) Euler's quadrilateral theorem states $${\displaystyle AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}+4x^{2},}$$ where ${\displaystyle x}$ is the length of the line segment joining the midpoints of the diagonals. It can be seen from the diagram that ${\displaystyle x=0}$ for a parallelogram, and so the general formula simplifies to the parallelogram law.

In the parallelogram on the right, let AD = BC = a, AB = DC = b, ${\displaystyle \angle BAD=\alpha .}$ By using the law of cosines in triangle ${\displaystyle \triangle BAD,}$ we get: $${\displaystyle a^{2}+b^{2}-2ab\cos(\alpha )=BD^{2}.}$$

In a parallelogram, adjacent angles are supplementary, therefore ${\displaystyle \angle ADC=180^{\circ }-\alpha .}$ Using the law of cosines in triangle ${\displaystyle \triangle ADC,}$ produces: $${\displaystyle a^{2}+b^{2}-2ab\cos(180^{\circ }-\alpha )=AC^{2}.}$$

By applying the trigonometric identity ${\displaystyle \cos(180^{\circ }-x)=-\cos x}$ to the former result proves: $${\displaystyle a^{2}+b^{2}+2ab\cos(\alpha )=AC^{2}.}$$

Now the sum of squares ${\displaystyle BD^{2}+AC^{2}}$ can be expressed as: $${\displaystyle BD^{2}+AC^{2}=a^{2}+b^{2}-2ab\cos(\alpha )+a^{2}+b^{2}+2ab\cos(\alpha ).}$$

Simplifying this expression, it becomes: $${\displaystyle BD^{2}+AC^{2}=2a^{2}+2b^{2}.}$$

In a normed space, the statement of the parallelogram law is an equation relating norms: $${\displaystyle 2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}\quad {\text{ for all }}x,y.}$$