PlanetMath: construct the center of a given circle

Draw any chord $\ol{AB}$ in the circle, and construct the perpendicular bisector of $\ol{AB}$ , intersecting $\ol{AB}$ 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 $\ol{DE}$ . Note that in the diagram below, $O$ is purposely drawn not to lie on $\ol{DE}$ ; the proof shows that this position is impossible and that in fact $O$ lies on $\ol{DE}$ . It then follows easily that in fact $O$ is the midpoint of $\ol{DE}$ .

\begin{pspicture*}(-2.8000,-2.8000)(2.6000,2.6000) \rput(-2.81,-2.81){.} \rput(2.61,2.61){.} \pscircle(0.0000,0.0000){2.0000} \psdots[dotstyle=*, dotscale=1.0000](-1.2856,-1.5321) \psdots[dotstyle=*, dotscale=1.0000](1.7321,-1.0000) \psline(-1.2856,-1.5321)(1.7321,-1.0000) \psline[linestyle=dotted](0.4585,-2.6000)(-0.4585,2.6000) \psdots[dotstyle=*, dotscale=1.0000](0.3473,-1.9696) \psdots[dotstyle=*, dotscale=1.0000](-0.3473,1.9696) \psline(0.3473,-1.9696)(-0.3473,1.9696) \psdots[dotstyle=*, dotscale=1.0000](0.2232,-1.2660) \psline(0.0017,-1.3051)(-0.0374,-1.0835)(0.1842,-1.0445) \psline(-0.5868,-1.2566)(-0.5347,-1.5520) \psline(-0.5277,-1.2461)(-0.4756,-1.5416) \psline(0.9221,-0.9905)(0.9741,-1.2860) \psline(0.9811,-0.9801)(1.0332,-1.2755) \uput{0.3000}[135.0000](0.2232,-1.2660){$C$} \uput{0.3000}[230.0000](-1.2856,-1.5321){$A$} \uput{0.3000}[330.0000](1.7321,-1.0000){$B$} \uput{0.3000}[315.0000](0.3473,-1.9696){$D$} \uput{0.3000}[45.0000](-0.3473,1.9696){$E$} \psdots[dotstyle=*, dotscale=1.0000](0.6000,0.5000) \uput{0.3000}[45.0000](0.6000,0.5000){$O$} \psline(0.6000,0.5000)(-1.2856,-1.5321) \psline(-0.2328,-0.6181)(-0.4527,-0.4140) \psline(0.6000,0.5000)(1.7321,-1.0000) \psline(1.2858,-0.1596)(1.0463,-0.3404) \psline(0.6000,0.5000)(0.2232,-1.2660) \end{pspicture*} \end{center} Since $O$ is the center of the circle, it follows that $OA=OB$. Since $\ol{DE}$ bisects $\ol{AB}$, we see in addition that $AC=BC$. $\triangle ACO$ and $\triangle BCO$ share their third \htmladdnormallink{side}{http://planetmath.org/encyclopedia/ExteriorAngle.html}, $\ol{OC}$. So by \htmladdnormallink{SSS}{http://planetmath.org/encyclopedia/AffineCongruence.html}, $\triangle ACO \cong \triangle BCO$, and thus, using \htmladdnormallink{CPCTC}{http://planetmath.org/encyclopedia/CPCTC.html}, $\angle ACO\cong\angle BCO$. But $\angle ACO+\angle BCO=180^{\circ}$, so $\angle ACO$ and $\angle BCO$ are each \htmladdnormallink{right angles}{http://planetmath.org/encyclopedia/Acute.html}. Thus $O$ in fact lies on $\ol{DE}$. However, since $O$ is the center of the circle, it must be equidistant from $D$ and $E$, and thus $O$ is the midpoint of $\ol{DE}$. \end{document}