Prove that an equilateral triangle can be constructed on any given line segment.

(N/A) In the statement above,a line segment of any length is given,say $AB$ [see Fig. $(i)$].
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. $(ii)$]. 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 $\Delta ABC$ [see Fig. $(iii)$].
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 circle $(1)$.
Similarly,$AB = BC$ (Radii of the same circle) $(2)$.
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,$\Delta ABC$ 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.