construct the center of a given circle

[Euclid, Book III, Prop. 1] Find the center (http://planetmath.org/Center8) 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 $\overline{AB}$ in the circle, and construct the perpendicular bisector of $\overline{AB}$, intersecting $\overline{AB}$ in $C$, and the circle in $D,E$.

Let $O$ be the center of the circle; we will show that $O$ is the midpoint of $\overline{DE}$. Note that in the diagram below, $O$ is purposely drawn not to lie on $\overline{DE}$; the proof shows that this position is impossible and that in fact $O$ lies on $\overline{DE}$. It then follows easily that in fact $O$ is the midpoint of $\overline{DE}$.

Since $O$ is the center of the circle, it follows that $OA=OB$. Since $\overline{DE}$ bisects $\overline{AB}$, we see in addition that $AC=BC$. $\mathrm{△}ACO$ and $\mathrm{△}BCO$ share their third side, $\overline{OC}$. So by SSS, $\mathrm{△}ACO\cong \mathrm{△}BCO$, and thus, using CPCTC, $\mathrm{\angle }ACO\cong \mathrm{\angle }BCO$. But $\mathrm{\angle }ACO+\mathrm{\angle }BCO={180}^{\circ }$, so $\mathrm{\angle }ACO$ and $\mathrm{\angle }BCO$ are each right angles. Thus $O$ in fact lies on $\overline{DE}$.

However, since $O$ is the center of the circle, it must be equidistant from $D$ and $E$, and thus $O$ is the midpoint of $\overline{DE}$.