Given the circle, midpoint $O(0,0)$ and radius $r$. On this circle, the equilateral $\triangle ABC\ :\ A(r,0),B(-\frac{r}{2},\frac{\sqrt{3}r}{2}),C(-\frac{r}{2},-\frac{\sqrt{3}r}{2})$. Choose $E(r\cos \alpha,r\sin \alpha)$. Equation of the pendicular bisector of $EO\ :\ y=\frac{1}{\sin \alpha}(\frac{r}{2}-x\cos \alpha)$. This line intersects the circle in the point $D(\frac{r(\cos \alpha+\sqrt{3}\sin \alpha)}{2},\frac{r(\sin \alpha-\sqrt{3}\cos \alpha)}{2}\ )$. Slope of the line $CE\ :\ m_{CE}=\frac{\sin \alpha+\frac{\sqrt{3}}{2}}{\cos \alpha+\frac{1}{2}}$, slope of the line $BD\ :\ m_{BD}=\frac{\sin \alpha-\sqrt{3}\cos \alpha-\sqrt{3}}{\cos \alpha+\sqrt{3}\sin \alpha+1}$. Because $m_{CE} \cdot m_{BD} + 1=0$, we have $CE \bot BD$.