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Proof: By Euclid

(related to Lemma: Lem. 10.016: Incommensurability of Sum of Incommensurable Magnitudes)

• For let the parallelogram $AD$, falling short by the square figure $DB$, have been applied to the straight line $AB$.
• I say that $AD$ is equal to the (rectangle contained) by $AC$ and $CB$.

• And it is immediately obvious.
• For since $DB$ is a square, $DC$ is equal to $CB$.
• And $AD$ is the (rectangle contained) by $AC$ and $CD$ - that is to say, by $AC$ and $CB$.
• Thus, if ... to some straight line, and so on ....
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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"