How to prove that the diagonals of a parallelogram are never parallel?

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.

• 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. – prashanth rao Feb 4 '18 at 15:27
• Can it be generalized for every quadrilateral? – dssknj Feb 5 '18 at 13:07

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.