Squares of Diagonals of Parallelogram

Theorem
Let $ABCD$ be a parallelogram.


 * DiameterOfParallelogram.png

Then:
 * $AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2$

Proof
But we have that $\angle C$ and $\angle B$ are supplementary:
 * $\angle C = \angle B = 180 \degrees$

Thus from Cosine of Supplementary Angle:
 * $\cos \angle B + \cos \angle C = 0$

The result follows.