(related to Proposition: Prop. 11.39: Prisms of Equal Height with Parallelogram and Triangle as Base)

- 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 $GHKLMN$.

- For let the solids $AO$ and $GP$ have been completed.
- Since parallelogram $AF$ is double triangle $GHK$, and parallelogram $HK$ is also double triangle $GHK$ [Prop. 1.34], parallelogram $AF$ is thus equal to parallelogram $HK$.
- And parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another [Prop. 11.31].
- Thus, solid $AO$ is equal to solid $GP$.
- And prism $ABCDEF$ is half of solid $AO$, and prism $GHKLMN$ half of solid $GP$ [Prop. 11.28].
- prism $ABCDEF$ is thus equal to prism $GHKLMN$.
- Thus, 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 are equal.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"