---
book: 11
number: 10
id: "XI.10"
kind: "theorem"
uses: ["[[book-1/proposition-33]]", "[[book-11/proposition-9]]", "[[book-1/proposition-8]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.10"
license: "CC-BY-SA-4.0"
---

# XI.10

*If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane*, *they will contain equal angles.*

## Proof

For let the two straight lines *AB*, *BC* meeting one another be parallel to the two straight lines *DE*, *EF* meeting one another, not in the same plane; I say that the angle *ABC* is equal to the angle *DEF*.

For let *BA*, *BC*, *ED*, *EF* be cut off equal to one another, and let *AD*, *CF*, *BE*, *AC*, *DF* be joined.

Now, since *BA* is equal and parallel to *ED*, therefore *AD* is also equal and parallel to *BE*. [[book-1/proposition-33|I. 33]]

For the same reason *CF* is also equal and parallel to *BE*.

Therefore each of the straight lines *AD*, *CF* is equal and parallel to *BE*.

But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [[book-11/proposition-9|XI. 9]] therefore *AD* is parallel and equal to *CF*.

And *AC*, *DF* join them; therefore *AC* is also equal and parallel to *DF*. [[book-1/proposition-33|I. 33]]

Now, since the two sides *AB*, *BC* are equal to the two sides *DE*, *EF*, and the base *AC* is equal to the base *DF*, therefore the angle *ABC* is equal to the angle *DEF*. [[book-1/proposition-8|I. 8]]

Therefore etc.