PROPOSITION 1.

On a given finite straight line[1] to construct an equilateral triangle.
C D A B E

Let AB be the given finite straight line.[2]

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;[3] [Post. 3]

again, with centre B and distance BA let the circle ACE be described; [Post. 3]

and from the point C,[4] in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1]

Now, since the point A is the centre of the circle CDB,

[Def. 15]AC is equal to AB.

Again, since the point B is the centre of the circle CAE,

[Def. 15]BC is equal to BA.

But CA was also proved equal to AB; therefore each of the straight lines CA, CB[5] is equal to AB.

And things which are equal to the same thing are also equal to one another; [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.

See Commentary on Proposition I.1.