Proposition I.1 · Problem
On a given finite straight line to construct an equilateral triangle.
Drag the points circled in red. The figure changes, and every step of the proof still holds.
- Let AB be the given finite straight line. It is required to construct an equilateral triangle on AB.
- With centre A and distance AB let the circle BCD be described.Postulate 3Postulate 3: To describe a circle with any centre and distance.
- Again, with centre B and distance BA let the circle ACE be described.Postulate 3Postulate 3: To describe a circle with any centre and distance.
- From the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined.Postulate 1Postulate 1: To draw a straight line from any point to any point.
- Now, since the point A is the centre of the circle CDB, AC is equal to AB.Definition 15Definition 15: A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another.
- Again, since the point B is the centre of the circle CAE, BC is equal to BA.Definition 15Definition 15: A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another.
- But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB. Things which are equal to the same thing are also equal to one another; therefore CA is also equal to CB. So the three straight lines CA, AB, BC are equal to one another, and the triangle ABC is equilateral, constructed on the given finite straight line AB.Common Notion 1Common Notion 1: Things which are equal to the same thing are also equal to one another.
Being what it was required to do.