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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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Sorry, this is utterly incomprehensible. Post it in some language in which you are comfortable, and someone will translate it into English for you. –  Gerry Myerson Sep 1 '12 at 6:13
    
Ok Gerry Myerson. I believe that now is correct and intelligible statement of the question. –  Elias Sep 1 '12 at 17:10
    
So what is the question? What is the $N$ foot? –  Thomas Sep 1 '12 at 17:12
    
Can you explain exactly about point M: it is midpoint for which arc ($\widehat{BC}$ long arc which containing point A )? Is it possible that point C can be between A and B? –  Mike Sep 2 '12 at 18:59
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Unlike @Gerry, I thought I was able to comprehend the first version of the question, but I cannot comprehend it now. –  Rahul Sep 2 '12 at 20:48
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1 Answer

up vote 4 down vote accepted

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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