---
book: 1
number: 2
id: "I.2"
kind: "construction"
uses: ["[[book-1/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.2"
license: "CC-BY-SA-4.0"
---

# I.2

To place at a given point (as an extremity) a straight line equal to a given straight line.

![I.2](figures/I-2.svg)

## Proof

Let *A* be the given point, and *BC* the given straight line.

Thus it is required to place at the point *A* (as an extremity) a straight line equal to the given straight line *BC*.

From the point *A* to the point *B* let the straight line *AB* be joined; [[book-1/postulates#Postulate 1|Post. 1]] and on it let the equilateral triangle *DAB* be constructed. [[book-1/proposition-1|I. 1]]

Let the straight lines *AE*, *BF* be produced in a straight line with *DA*, *DB*; [[book-1/postulates#Postulate 2|Post. 2]] with centre *B* and distance *BC* let the circle *CGH* be described; [[book-1/postulates#Postulate 3|Post. 3]] and again, with centre *D* and distance *DG* let the circle *GKL* be described. [[book-1/postulates#Postulate 3|Post. 3]]

Then, since the point *B* is the centre of the circle *CGH*, *BC* is equal to *BG*.

Again, since the point *D* is the centre of the circle *GKL*, *DL* is equal to *DG*.

And in these *DA* is equal to *DB*; therefore the remainder *AL* is equal to the remainder *BG.* [[book-1/common-notions#Common Notion 3|C.N. 3]]

But *BC* was also proved equal to *BG*; therefore each of the straight lines *AL*, *BC* is equal to *BG*.

And things which are equal to the same thing are also equal to one another; [[book-1/common-notions#Common Notion 1|C.N. 1]] therefore *AL* is also equal to *BC*.

Therefore at the given point *A* the straight line *AL* is placed equal to the given straight line *BC*.

(Being) what it was required to do.