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In following, ($x$) denotes condition $x$, and ($x$') denotes condition $\neg x$.

For a polygon of $2n$ sides, let it be given that (a) opposite sides are parallel. It is not difficult to find examples where (b') opposite interior angles differ and (c') opposite sides are of different lengths.

Example: Octagon (1,0), (2,1), (0,1), (0,2), (3,2), (4,3), (5,3), (5,0), (1,0).

I am tempted to conjecture that if the polygon is convex and opposite sides are parallel, then (b) opposite interior angles match and (c) opposite sides have equal length. But I don't see how to prove it and don't have a counterexample.

  1. Does anyone know of a counterexample? Update 1: Following the first remark by André Nicolas, here's a construction of convex polygon Q where (a) and (b) hold but not (c): Let P be a regular hexagon, with one edge extending from (-5,0) to (+5,0). Intersect P with all points where -6 =< x =< 7 to form an octagon Q.

  2. If the conjecture is false, what conditions would be sufficient to enforce (b) or (c) ? (It seems obvious that (a) and (b) imply (c), while (a) and (c) imply (b))

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    ( Convex + (a) ) $\implies$ (b): draw the diagonal and consider the "Z"s (or are they "N"s?) made with the sides.2011-09-10

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For (c), consider a non-square rectangle. For (b) if you define opposite sides as half the sides around the polygon, you should have them equal by the theorem that corresponding angles with a transversal of parallel lines are equal.

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    We know that a parallelogram's opposite angles are equal because they are both supplementary to the same angle (either one of the other angles of the parallelogram). Still no diagonal necessary.2011-09-10