Squares of Diagonals of Parallelogram
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Theorem
Let $ABCD$ be a parallelogram.
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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $127$