Proposition: Prop. 11.39: Prisms of Equal Height with Parallelogram and Triangle as Base

(Proposition 39 from Book 11 of Euclid's “Elements”)

If there are two equal height prisms, and one has a parallelogram, and the other a triangle, (as a) base, and the parallelogram is double the triangle, then the prisms will be equal. * Let $ABCDEF$ and $GHKLMN$ be two equal height prisms, and let the former have the parallelogram $AF$, and the latter the triangle $GHK$, as a base. * And let parallelogram $AF$ be twice triangle $GHK$. * I say that prism $ABCDEF$ is equal to prism $GHKLMN$.


Modern Formulation

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Proofs: 1

Proofs: 1 2

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Adapted from (subject to copyright, with kind permission)

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

Adapted from CC BY-SA 3.0 Sources:

  1. Prime.mover and others: "Pr∞fWiki",, 2016