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Proposition: 2.05: Rectangle is Difference of Two Squares

(Proposition 5 from Book 2 of Euclid's “Elements”)

If a straight line is cut into equal and unequal (pieces) then the rectangle contained by the unequal pieces of the whole (straight line), plus the square on the (difference) between the (equal and unequal) pieces, is equal to the square on half (of the straight line). * For let any straight line $AB$ have been cut - equally at $C$, and unequally at $D$. * I say that the rectangle contained by $AD$ and $DB$, plus the square on $CD$, is equal to the square on $CB$.

Modern Formulation

With \(a:=AC=CB\) and \(b:=CD\), this propositions states that \[(a+b)(a-b)=a^2-b^2\]

which is one of the binomial formulae.

Table of Contents

Proofs: 1

Mentioned in:

Proofs: 1 2 3 4 5
Sections: 6

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:

@Calahan

@Casey

@Fitzpatrick

References

Adapted from CC BY-SA 3.0 Sources:

1. Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014

Adapted from (Public Domain)

1. Casey, John: "The First Six Books of the Elements of Euclid"

Adapted from (subject to copyright, with kind permission)

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