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Proposition: 1.41: Parallelograms and Triagles

(Proposition 41 from Book 1 of Euclid's “Elements”)

If a parallelogram has the same base as a triangle, and is between the same parallels, then the parallelogram is double (the area) of the triangle. * For let parallelogram $ABCD$ have the same base $BC$ as triangle $EBC$, and let it be between the same parallels, $BC$ and $AE$. * I say that parallelogram $ABCD$ is double (the area) of triangle $BEC$.

Modern Formulation

If a parallelogram (\(\boxdot{ABCD}\)) and a triangle (\(\triangle{EBC}\)) stand on the same base (\(\overline{BC}\)) and between the same parallels (\(\overline{AE}\), \(\overline{BC}\)), then the parallelogram is double the area of the triangle.

Table of Contents

Proofs: 1

Mentioned in:

Proofs: 1 2 3

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

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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"