Solution : In the statement above, a line segment of any length is given, say AB [see Fig. 5.8(i)].Here, you need to do some construction. Using Euclid's Postulate 3, you can draw a circle with point A as the centre and AB as the radius [see Fig. 5.8(ii)]. Similarly, draw another circle with point B as the centre and BA as the radius. The two circles meet at a point, say C. Now, draw the line segments AC and BC to form △ABC [see Fig. 5.8 (iii)].So, you have to prove that this triangle is equilateral, i.e., AB=AC=BC.Now, AB=AC, since they are the radii of the same circleSimilarly, AB=BC (Radii of the same circle)From these two facts, and Euclid's axiom that things which are equal to the same thing are equal to one another, you can conclude that AB=BC=AC.So, △ABC is an equilateral triangle.Note that here Euclid has assumed, without mentioning anywhere, that the two circles drawn with centres A and B will meet each other at a point.