PlanetMath: opposing angles in a cyclic quadrilateral are supplementary

Proof. Let $ABCD$ be a quadrilateral inscribed in a circle

\begin{pspicture*}(-3.0000,-3.0000)(3.0000,3.0000) \rput(-3.1,-3.1){.} \rput(3.1,3.1){.} \psdots[dotstyle=*, dotscale=1.0000](0.0000,0.0000) \uput{0.3000}[45.0000](0.0000,0.0000){$O$} \pscircle(0.0000,0.0000){2.0000} \psdots[dotstyle=*, dotscale=1.0000](2.0000,0.0000) \uput{0.3000}[0.0000](2.0000,0.0000){$A$} \psdots[dotstyle=*, dotscale=1.0000](0.5176,1.9319) \uput{0.3000}[75.0000](0.5176,1.9319){$B$} \psdots[dotstyle=*, dotscale=1.0000](-1.9696,-0.3473) \uput{0.3000}[190.0000](-1.9696,-0.3473){$C$} \psdots[dotstyle=*, dotscale=1.0000](1.5321,-1.2856) \uput{0.3000}[320.0000](1.5321,-1.2856){$D$} \pspolygon(2.0000,0.0000)(0.5176,1.9319)(-1.9696,-0.3473)(1.5321,-1.2856) \psline(0.0000,0.0000)(0.5176,1.9319) \psline(0.0000,0.0000)(1.5321,-1.2856) \end{pspicture*} \end{center} Note that $\angle BAD$ subtends \htmladdnormallink{arc}{http://planetmath.org/encyclopedia/Arc2.html} $BCD$ and $\angle BCD$ subtends arc $BAD$. Now, since a \htmladdnormallink{circumferential angle is half the corresponding central angle}{http://planetmath.org/encyclopedia/CircumferentialAngle.html}, we see that $\angle BAD + \angle BCD$ is one half of the sum of the two angles $BOD$ at $O$. But the sum of these two angles is $360^{\circ}$, so that$$\angle BAD + \angle BCD = 180^{\circ$$ Similarly, the sum of the other two opposing angles is also $180^{\circ}$. \end{proof} \end{document}

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