I am trying to show that in hyperbolic geometry, any diagonal of a parallellogram $\square ABCD$ divides the parallellogram into two non-congruent triangles.
I have tried assuming the contrary, and used that the summit of a Saccheri quadrilateral is longer than the base, and that the in a Lambert quadrialteral, the length of a side is less than that of the opposite side, but I have not been able to derive a contradiction despite many efforts. Another possibility would be to construct a triangle based on this having angle sum $\geq 180$, or showing that some linear pair of angles violate the Linear Pair Theorem.
Can anyone give me a hint on how to approach this problem?
I have been able to show that it cannot be true for every parallellogram $\square ABCD$, that is; I have been able to show that there exists a parallellogram $\square ABCD$ having a diagonal that divides it into noncongruent triangles. I did that as follows:
Let $l=\overleftrightarrow{AB}$, and let $E \neq A$ be a point of intersection between $l$ and some line $m$ through $D$ parallell to $\overleftrightarrow{BC}$ (so $m \neq \overleftrightarrow{AD}$). Then $\square EBCD$ is also a parallellogram. If the the diagonals of this parallellogram divides it into congruent triangles, then $EB=DC=AB$, giving the contradiction $EA=0$.
But this does not prove that the statement is true for every parallellogram....
For a counterexample, take a parallelogram which is a regular quadrilateral.