To bisect a given finite straight line.

Let *AB* be the given finite straight line.

It is required to bisect the finite straight line *AB.*

Construct the equilateral triangle *ABC* on it, and bisect the angle *ACB* by the straight line *CD.*

I say that the straight line *AB* is bisected at the point *D.*

Since *CA* equals *CB,* and *CD* is common, therefore the two sides *CA* and *CD* equal the two sides *CB* and *CD* respectively, and the angle *ACD* equals the angle *BCD,* therefore the base *AD* equals the base *BD.*

Therefore the given straight line *AB* is bisected at *D.*

Q.E.F.

First, the equilateral triangle ABC needs to be constructed. According to I.1 two circles need to be drawn: one with center A and radius AB, the other with center B and radius BA. One of the points of intersection of the two circles is C. Then to bisect angle ACB, according to I.9, an arbitrary point is chosen on one side of the angle, and it might as well be the point A on the side AC, and a point equally far from C on the side BC, which is, of course, B. Then an equilateral triangle is constructed on the line AB. There are two such equilateral triangles, the one already constructed ACB, and another one, call it AEB. The point E is the other intersection of the two circles already drawn. Then, by I.9, the line CE bisects the angle ACB, and according to this proposition, the point D bisects the line AB.
Actually, only two circles and the straight line |