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How to rigorously prove that the diagonals of a parallelogram are never parallel? It is intuitively obvious, but since it is not an axiom, it is a proposition that needs to be proved. I would like to see a proof without using analytic geometry, but only the old methods of Euclidean synthetic geometry.

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    $\begingroup$ The sum of the internal angles within a pair of parallel sides of the parallelogram is 180 degrees. Therefore the sum of internal angles corresponding to the two diagonals must be less than 180, which implies that the two diagonals must intersect with one another. $\endgroup$ – prashanth rao Feb 4 '18 at 15:27
  • $\begingroup$ Can it be generalized for every quadrilateral? $\endgroup$ – dssknj Feb 5 '18 at 13:07
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The diagonals of a parallelogram bisect each other. That is, they intersect each other at their midpoints, and so can not be parallel. A proof of the bisection property is based on the two families of lines parallel to the sides of the parallelogram. Any member of one family intersect any member of the other family. Given any pair of opposite sides of the parallelogram, joining the midpoints of the opposite sides gives a member of the opposite family. Do this for the other pair of opposite sides and the two lines intersect in the midpoint of the parallelogram. This point is also the midpoint of the two diagonals.

By the way, notice that being a parallelogram is an affine property, and so is being a midpoint of a line segment. Also, every parallelogram is affinely equivalent to a square. Thus, it is enough to prove that the diagonals of any square bisect each other.

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  • $\begingroup$ I think you should eliminate the reference to the Wikipedia proof, as it begins by assuming the diagonals intersect. $\endgroup$ – fredgoodman Feb 4 '18 at 23:22
  • $\begingroup$ @fredgoodman. You are right. Removed the reference. $\endgroup$ – Somos Feb 5 '18 at 1:09
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    $\begingroup$ I think that if one traces back the proof of the "proportional segments" theorem regarding transversals to 3 parallel lines, one will eventually find some dependence on Pasch's axiom. I wrote down another proof that uses Pasch's lemma explicitly, but never mind that. It seems to me that instead, one might as well prove, synthetically, that the diagonals of a convex quadrilateral intersect in the interior of the quadrilateral, using Pasch-like statements about lines and triangles. $\endgroup$ – fredgoodman Feb 5 '18 at 2:26
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Let ABCD be a parallelogram, AC and BD are its diagonals. Assume that $AC\mid\mid BD$. Now, $AB\mid\mid CD$ and $AC\mid\mid BD.$ Now we can move from A to B to D to C to A, without intersecting any other line at points other than A, B, C, D. Here we have got an another parallelogram named ABDC, which is not possible since 4 points define a unique quadrilateral (convex) . Hence, our assumption is wrong AC can't be parallel to BD.

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