|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.10
|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.||I.Def.15
|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.||I.Def.10|
|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.
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.
The corollary is used in propositions III.9 and III.10.
Next proposition: III.2
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