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Consider a polygon $A_{0} \dots A_{n}$. Why does there exist diagonal $A_{i}A_{j}$ without intersection of polygon. It's looks obvious but why?

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  • $\begingroup$ First of all, don't put the word "stupid" in your title. Second, your question is unclear. What do you mean by "...without intersection of other polygon"? Lastly, what have you tried yourself? $\endgroup$ Apr 24 '17 at 9:12
  • $\begingroup$ @vrugtehagel my bad , sorry $\endgroup$
    – openspace
    Apr 24 '17 at 9:13
  • $\begingroup$ It does hold for concave polygons, if I understand the OP correctly $\endgroup$ Apr 24 '17 at 9:15
  • $\begingroup$ @vrugtehagel Thanks for the reminder. I took existence as "for all"... $\endgroup$
    – Frenzy Li
    Apr 24 '17 at 9:18
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A convex polygon is a non-empty intersection of convex sets (half-planes), hence it is convex. In particular, if we consider the segment joining $A_i$ and $A_j$ with $|i-j|>1$, such segment lies inside the polygon, since it is the convex envelope of $A_i$ and $A_j$.

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