On a given finite straight line to construct an equilateral triangle.
Proof
Let AB be the given finite straight line. With centre A and
distance AB describe the circle BCD (Postulate 3). With centre
B and distance BA describe the circle ACE (Postulate 3). From
the point C, where the circles cut one another, draw CA and CB
(Postulate 1). Since A is the centre of BCD, AC=AB
(Definition I.15). Since B is the centre of ACE, BC=BA
(Definition I.15). By Common Notion 1, AC=BC. Therefore the
triangle ABC is equilateral.
Figure
Proposition I.1. The two circles with centres A and B and common radius AB meet at C; the triangle ABC is equilateral.