---
book: 1
number: 22
id: "I.22"
kind: "construction"
uses: ["[[book-1/proposition-20]]", "[[book-1/proposition-3]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.22"
license: "CC-BY-SA-4.0"
---

# I.22

Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [[book-1/proposition-20|I. 20]]

![I.22](figures/I-22.svg)

## Proof

Let the three given straight lines be *A*, *B*, *C*, and of these let two taken together in any manner be greater than the remaining one, namely *A*, *B* greater than *C*, *A*, *C* greater than *B*, and *B*, *C* greater than *A*; thus it is required to construct a triangle out of straight lines equal to *A*, *B*, *C*.

Let there be set out a straight line *DE*, terminated at *D* but of infinite length in the direction of *E*, and let *DF* be made equal to *A*, *FG* equal to *B*, and *GH* equal to *C*. [[book-1/proposition-3|I. 3]]

With centre *F* and distance *FD* let the circle *DKL* be described; again, with centre *G* and distance *GH* let the circle *KLH* be described; and let *KF*, *KG* be joined;

I say that the triangle *KFG* has been constructed out of three straight lines equal to *A*, *B*, *C*.

For, since the point *F* is the centre of the circle *DKL*, *FD* is equal to *FK*.

But *FD* is equal to *A*; therefore *KF* is also equal to *A*.

Again, since the point *G* is the centre of the circle *LKH*, *GH* is equal to *GK*.

But *GH* is equal to *C*; therefore *KG* is also equal to *C*. And *FG* is also equal to *B*; therefore the three straight lines *KF*, *FG*, *GK* are equal to the three straight lines *A*, *B*, *C*.

Therefore out of the three straight lines *KF*, *FG*, *GK*, which are equal to the three given straight lines *A*, *B*, *C*, the triangle *KFG* has been constructed.

Q. E. F.