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[Euclid, Book III, Prop. 1] Find the center of a given circle.
Since, in Euclidean geometry, a circle has one center only, it suffices to construct a point that is a center of the given circle.
Draw any chord
 in the circle, and construct the perpendicular bisector of
 , intersecting
 in  , and the circle in  .
Let  be the center of the circle; we will show that  is the midpoint of
 . Note that in the diagram below,  is purposely drawn not to lie on
 ; the proof shows that this position is impossible and that in fact  lies on
 . It then follows easily that in fact  is the midpoint of
 .
Since  is the center of the circle, it follows that  . Since
 bisects
 , we see in addition that  .
 and
 share their third side,
 . So by SSS,
 , and thus, using CPCTC,
 . But
 , so
 and
 are each right angles. Thus  in fact lies on
 .
However, since  is the center of the circle, it must be equidistant from  and  , and thus  is the midpoint of
 .
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