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How to prove that each non-convex polygon with no self-intersecting parts, has at least one interior angle which size is less then $180$ degrees.

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In case of self-intersecting polygons, what is your definition of "interior angle"? –  dtldarek Jan 25 '13 at 11:35
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2 Answers

up vote 3 down vote accepted

Hint: Consider the vertex with the largest $x$-coordinate.

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The formula for the sum of the angles of a polygon, as a function of the number of sides, holds for non-convex as well as for convex polygons, and the result follows immediately from that formula.

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Can you please share proof of that formula for non-convex polygons –  Ashot Jan 25 '13 at 11:54
    
Why not first look at the proof for convex polygons, and see whether it works? –  Gerry Myerson Jan 25 '13 at 12:25
    
There the polygon is divided into triangles, which is not trivial for non-convex polygons, in fact I need to prove it using this affirmation –  Ashot Jan 25 '13 at 12:38
    
That's one way. Another is by considering "exterior angles". –  Gerry Myerson Jan 25 '13 at 23:01
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