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
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References
Adapted from CC BYSA 3.0 Sources:
 Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014
Adapted from (Public Domain)
 Casey, John: "The First Six Books of the Elements of Euclid"
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"