---
book: 11
number: 22
id: "XI.22"
kind: "construction"
uses: ["[[book-1/proposition-4]]", "[[book-1/proposition-24]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.22"
license: "CC-BY-SA-4.0"
---

# XI.22

*If there be three plane angles of which two*, *taken together in any manner*, *are greater than the remaining one*, *and they are contained by equal straight lines*, *it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.*

## Proof

Let there be three plane angles *ABC*, *DEF*, *GHK*, of which two, taken together in any manner, are greater than the remaining one, namely the angles *ABC*, *DEF* greater than the angle *GHK*, the angles *DEF*, *GHK* greater than the angle *ABC*, and, further, the angles *GHK*, *ABC* greater than the angle *DEF*; let the straight lines *AB*, *BC*, *DE*, *EF*, *GH*, *HK* be equal, and let *AC*, *DF*, *GK* be joined; I say that it is possible to construct a triangle out of straight lines equal to *AC*, *DF*, *GK*, that is, that any two of the straight lines *AC*, *DF*, *GK* are greater than the remaining one.

Now, if the angles *ABC*, *DEF*, *GHK* are equal to one another, it is manifest that, *AC*, *DF*, *GK* being equal also, it is possible to construct a triangle out of straight lines equal to *AC*, *DF*, *GK*.

But, if not, let them be unequal, and on the straight line *HK*, and at the point *H* on it, let the angle *KHL* be constructed equal to the angle *ABC*; let *HL* be made equal to one of the straight lines *AB*, *BC*, *DE*, *EF*, *GH*, *HK*, and let *KL*, *GL* be joined.

Now, since the two sides *AB*, *BC* are equal to the two sides *KH*, *HL*, and the angle at *B* is equal to the angle *KHL*, therefore the base *AC* is equal to the base *KL*. [[book-1/proposition-4|I. 4]]

And, since the angles *ABC*, *GHK* are greater than the angle *DEF*, while the angle *ABC* is equal to the angle *KHL*, therefore the angle *GHL* is greater than the angle *DEF*.

And, since the two sides *GH*, *HL* are equal to the two sides *DE*, *EF*, and the angle *GHL* is greater than the angle *DEF*, therefore the base *GL* is greater than the base *DF*. [[book-1/proposition-24|I. 24]]

But *GK*, *KL* are greater than *GL*.

Therefore *GK*, *KL* are much greater than *DF*.

But *KL* is equal to *AC*; therefore *AC*, *GK* are greater than the remaining straight line *DF*.

Similarly we can prove that *AC*, *DF* are greater than *GK*, and further *DF*, *GK* are greater than *AC*.

Therefore it is possible to construct a triangle out of straight lines equal to *AC*, *DF*, *GK*. Q. E. D.