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Three points $A$, $B$ and $C$ are on a circle, $G$. Suppose $\overline{AB}>\overline{AC}$. Let $M$ be the midpoint of the arc of the circle containing the points A and N the point in $AB$ such that $MN \perp AB$.

I would like to prove that $\overline{AC}=\overline{AN}+\overline{NB}$.

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    As \overline{AB}>\overline{AC} then $\overline{AC}=\overline{AN}+\overline{NB}=\overline{AB}$ not true. So, I think @Rahul Narain is right in his guess.2012-09-04

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If Rahul Narain is right in his vision of you question: it is simple solution to prove that $|BN|=|AN|+|AC|$.

$MG$ - line through the center $O$ of circle. $HG||MN \Rightarrow NIFG$ - rectangle and $|NI|=|FG|$. $|FG|=|AC|$ because of $\angle ABC = \angle FMG$. $|AN|=|IB|$ because $HG||MN$ and both lines are on equal distance from $O$. So we have that $|AC|+|AN|=|NI|+|IB|=|BN|$.

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