Show that the chord $DE$ bisects the chord $BC$ in a circle Suppose a circle is centered at $O$, $CD$ is a chord perpendicular to a diameter $AB$, and a chord $AE$ bisects the radius $OC$. Show that the chord $DE$ bisects the chord $BC$.

If $N$ is the middle point of $CB$, $MN$ is a middle line in $OBC$, so $MN \parallel OB$. But I can't figure out how to prove this. Can you help me, please? Thanks!

Let $\{F\} = \overline{AB} \cap \overline{CD}$. Note that $\angle AED = \angle ACD$ by the Inscribed Angle Theorem, and also $\triangle ACF \sim \triangle ABC$, so $\angle ACD = \angle ABC = \angle OCB$. Therefore, $\angle OCB = \angle AED$.
In other words, $\angle MEN = \angle MCN$, so $MCEN$ is cyclic. From that, $\color{blue}{\angle EMN} = \angle ECN = \angle ECB = \color{blue}{\angle EAB}$, so $MN \parallel AB$. 
Since $M$ bisects $\overline{OC}$, it follows that $N$ bisects $\overline{BC}$. 
