Let AB be the given finite straight line.

Thus it is required to bisect the given straight line AB.

Let the equilateral triangle ABC be constructed on it, [I. 1]
and let the angle ACB be bisected by the straight line CD; [I. 9]
I say that the straight line AB has been bisected at the point D.

For, since AC is equal to CB, and CD is common,
the two sides AC, CD are equal to the two sides BC, CD respectively;
and the angle ACD is equal to the angle BCD;
therefore the base AD is equal to the base BD; [I. 4]

Therefore the given finite straight line AB has been bisected at D. Q.E.F.