Proposition 1

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.

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I.10
I.11, I.10

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.

I.Def.15
I.8

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.

I.Def.10

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.

Corollary

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.

Guide

 Since the definition of a circle, I.Def.15, includes the existence of a center, Euclid is justified in taking a point G as the center.

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.

Use of this proposition and its corollary

About half the proofs in Book III and several of those in Book IV begin with taking the center of a given circle, but in plane geometry, it isn’t necessary to invoke this proposition III.1 since the only way that circles can occur is if they were constructed around a center to begin with. Even in solid geometry, the center of a circle is usually known so that III.1 isn’t necessary. Indeed, that is the case whenever the center is needed in Euclid’s books on solid geometry (see XI.23, XIII.9 through XIII.13, and XIII.16). Sections of spheres cut by planes are also circles as are certain plane sections of cylinders and cones, and as the spheres, cylinders, and cones were generated by rotating semicircles, rectangles, and triangles about their sides, the center of the circle is known to be at the intersection of the side and the plane. In that sense, this proposition is redundant.

The corollary is used in propositions III.9 and III.10.