Squares of Diagonals of Parallelogram

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Theorem

Let $ABCD$ be a parallelogram.

DiameterOfParallelogram.png

Then:

$AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2$


Proof

\(\ds AC^2\) \(=\) \(\ds AB^2 + BC^2 - 2 (AB) (BC) \cos \angle B\) Cosine Rule
\(\ds BD^2\) \(=\) \(\ds BC^2 + CD^2 - 2 (CD) (BC) \cos \angle C\) Cosine Rule
\(\ds \) \(=\) \(\ds DA^2 + CD^2 - 2 (AB) (CD) \cos \angle C\) as $AB = CD$ and $BC = DA$
\(\ds \leadsto \ \ \) \(\ds AC^2 + BD^2\) \(=\) \(\ds AB^2 + BC^2 + DA^2 + CD^2 - 2 (AB) (BC) \paren {\cos \angle B + \cos \angle C}\)

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.

$\blacksquare$


Sources