document.write( "Question 843884: If a polygon has n sides, how many triangles are formed by drawing all diagonals from one vertex?
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Algebra.Com's Answer #508409 by KMST(5328)\"\" \"About 
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I would make that polygon a convex polygon,
\n" ); document.write( "like this: so all diagonals will be inside the polygon, and no triangles will overlap.
\n" ); document.write( "The answer may be the same for a concave polygon, like this
\n" ); document.write( " , but it is harder to visualize with concave polygons.
\n" ); document.write( "I believe the answer is \"highlight%28n-2%29\"
\n" ); document.write( "A polygon with \"n\" sides has \"n\" vertices.
\n" ); document.write( "From one of the n vertices, you can draw \"n-3\" diagonals to all \"n-1\" other vertices except the \"2\" adjacent ones.
\n" ); document.write( "(The lines connecting to the \"2\" adjacent vertices are sides of the polygon, and we cannot call them diagonals).
\n" ); document.write( "To one side of each diagonal is a triangle, and you count \"n-3\" of those:
\n" ); document.write( "one to that side of the first diagonal,
\n" ); document.write( "a second one to that side of the second diagonal, and so on.
\n" ); document.write( "You count \"n-3\" triangles that way.
\n" ); document.write( "There is \"1\" more triangle to the other side of the last of those diagonals.
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