To find the center of a given circle.

Let *ABC* be the given circle.

It is required to find the center of the circle *ABC.*

Draw a straight line *AB* through it at random, and bisect it at the point *D.* Draw *DC* from *D* at right angles to *AB,* and draw it through to *E.* Bisect *CE* at *F.*

I say that *F* is the center of the circle *ABC.*

For suppose it is not, but, if possible, let *G* be the center. Join *GA, GD,* and *GB.*

Then, since *AD* equals *DB,* and *DG* is common, the two sides *AD* and *DG* equal the two sides *BD* and *DG* respectively. And the base *GA* equals the base *GB,* for they are radii, therefore the angle *ADG* equals the angle *GDB.*

But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, therefore the angle *GDB* is right.

But the angle *FDB* is also right, therefore the angle *FDB* equals the angle *GDB,* the greater equals the less, which is impossible. Therefore *G* is not the center of the circle *ABC.*

Similarly we can prove that neither is any other point except *F.*

Therefore the point *F* is the center of the circle *ABC.*

Q.E.F.

From this it is clear that if in a circle a straight line cuts a straight line into two equal parts and at right angles, then the center of the circle lies on the cutting straight line.

In this proof *G* is shown to lie on the perpendicular bisector of the line *AB.* He leaves to the reader to show that *G* actually is the point *F* on the perpendicular bisector, but that’s clear since only the midpoint *F* is equidistant from the two points *C* and *E* on the circle. From that observation it also follows that the center of a circle is unique, although the uniqueness can easily be proved in other ways.

As Todhunter remarked, Euclid implicitly assumes that the perpendicular bisector of *AB* actually intersects the circle in points *C* and *E.*