# Is it always possible to cross two opposite pairs of adjacent sides of a convex 2n-gon by two parallel lines, not crossing through vertex?

We are given a convex polygon with even number of vertexes ($$ABCDEF$$ in my picture). We want to prove that it is always possible to find two pairs of adjacent sides that have the same amount of vertexes between them if we move from one adjacent pair of sides to another clockwise and counterclockwise, we call them “opposite” pairs, such that, we can draw two parallel lines that cross these opposite pairs of sides in their inner parts (no crossing in vertexes allowed), one line crosses one adjacent pair and another line crosses another pair. In my picture in pentagon $$ABCDEF$$ we have three possible opposite pairs with common vertexes in $$A$$ and $$D$$, $$B$$ and $$E$$, $$C$$ and $$F$$. Out of these three pairs only one (with vertexes in $$A$$ and $$D$$) allows such crossing with parallel lines $$L1$$ and $$L2$$, other two pairs cannot be crossed by two parallel lines because $$BC||DE$$ and $$CD||EF$$. Construction similar to the one in my picture, where one vertex of any 2D zonotope with even number of vertexes (vertex $$A’$$ of parallelogram $$A’CDE$$) is cut out like in my picture, creating three new vertexes $$A, B, F$$, proves that there exist convex 2n-gons with arbitrary large amount of vertexes, where there is only one possibility to cross two opposite pairs of adjacent sides with parallel lines, namely the newly created pair ($$FA$$ and $$AB$$ in my picture) and opposite pair ($$CD$$ and $$DE$$ in my picture), all other $$n-1$$ pairs cannot be crossed because of parallel sides property of original zonotope. But this only possible pair for converted zonotope case will always allow crossing, and in general we can find at least one such pair in any 2n-gon.

To prove above statement, we will use the following lemma –

Given division of interval [0;1] into 2n consecutive subintervals of nonzero length, each smaller than ½ in length, $$Int^1, …, Int^{2n}$$, we can always find a pair of intervals with index difference n (like $$Int^1$$ and $$Int^{n+1}$$), such, that they contain in their inner parts two points, one point inside each interval, at distance ½.

My proof of this Lemma is found here: Interval [0;1] is divided into 2*n sub-intervals, each smaller than ½, find two intervals with index difference n, containing points with distance ½

If we use interval $$2 \pi$$ instead of $$[0;1]$$ used in Lemma, convert angles between adjacent sides of convex 2n-gon ($$a, b, c, d, e, f$$ in my picture) into consecutive sub-intervals in this Lemma like this - $$Int^1 = (0;a), Int^2 = (a;a+b), Int^3 = (a+b;a+b+c),…$$ and note that two opposite pairs of sides can be crossed by two parallel lines iff corresponding sub-intervals in Lemma have in their inner parts points at distance $$\pi$$, we prove that such opposite pairs of sides exist for any convex 2n-gon.