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Proposition: 1.02: Constructing a Segment Equal to an Arbitrary Segment

(Proposition 2 from Book 1 of Euclid's “Elements”)

To place a straight line equal to a given straight line at a given point (as an extremity).

• Let $A$ be the given point, and $BC$ the given straight line.
• So it is required to place a straight line at point $A$ equal to the given straight line $BC$.

Modern Formulation

Given an arbitrary point \(A\) and an arbitrary segment \(\overline{BC}\), it is possible to construct a segment \(\overline{AF}\) such that its length is equal to the length of \(\overline{BC}\).

Table of Contents

Proofs: 1

Mentioned in:

Proofs: 1
Sections: 2

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

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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"