Problem. On a given finite right line (AB) to construct an equilateral triangle.
Sol.—With A as centre, and AB as radius, describe the circle BCD (Post. iii.). With B as centre, and BA as radius, describe the circle ACE, cutting the former circle in C. Join CA, CB (Post. i.). Then ABC is the equilateral triangle required.
Dem.—Because A is the centre of the circle BCD, AC is equal to AB (Def. xxxii.). Again, because B is the centre of the circle ACE, BC is equal to BA. Hence we have proved.
AC = AB, and BC = AB. But things which are equal to the same are equal to one another (Axiom i.); therefore AC is equal to BC; therefore the three lines AB, BC, CA are equal to one another. Hence the triangle ABC is equilateral (Def. xxi.); and it is described on the given line AB, which was required to be done.