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

# I.1

On a given finite straight line to construct an equilateral triangle.

![I.1](figures/I-1.svg)

## Proof

Let *AB* be the given finite straight line.

Thus it is required to construct an equilateral triangle on the straight line *AB*.

With centre *A* and distance *AB* let the circle *BCD* be described; [[book-1/postulates#Postulate 3|Post. 3]] again, with centre *B* and distance *BA* let the circle *ACE* be described; [[book-1/postulates#Postulate 3|Post. 3]] and from the point *C*, in which the circles cut one another, to the points *A*, *B* let the straight lines *CA*, *CB* be joined. [[book-1/postulates#Postulate 1|Post. 1]]

Now, since the point *A* is the centre of the circle *CDB*, *AC* is equal to *AB*. [[book-1/definitions#Definition 15|Def. 15]]

Again, since the point *B* is the centre of the circle *CAE*, *BC* is equal to *BA*. [[book-1/definitions#Definition 15|Def. 15]]

But *CA* was also proved equal to *AB*; therefore each of the straight lines *CA*, *CB* is equal to *AB*.

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 *CA* is also equal to *CB*.

Therefore the three straight lines *CA*, *AB*, *BC* are equal to one another.

Therefore the triangle *ABC* is equilateral; and it has been constructed on the given finite straight line *AB*.

(Being) what it was required to do.