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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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