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I am not sure if the following result is well known. I stumbled across it from the paper The Perimetric Bisection of Triangles by Dov Avishalom, where the result was stated without proof. I am looking for a simple proof of the following

The following figure depicts a circular arc with chord $\mathrm{LN}$. The point $\mathrm{P}$ denotes the midpoint of the circular arc. We drop perpendicular from $\mathrm{P}$ to $\mathrm{LM}$, intersecting it at point $\mathrm{Q}$.

Then the claim is that $\mathrm{Q}$ bisects the broken line segment $\mathrm{LMN}$, that is we have $\mathrm{LQ} = \mathrm{QM}+\mathrm{MN}$.

circular arc

Thanks for any help.

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up vote 1 down vote accepted

This is the famous Broken Chord Theorem that goes back to Archimedes. With the right name to search for, you will easily find proofs.

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Ah, so it is a well known result. Thank you André. – EuYu Sep 16 '12 at 5:44

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