Suppose that we are given side lengths $a, b, c, d$ of a trapezoid. We know that two of them are parallel, and all values are different.
Moreover, we are given the height $h$ from the base (distance between two parallel lines).
The task is to find which sides are parallel.
I tried to form a triangle to use the triangle similarity, but I don't know how to proceed from there.
Assuming that the diagonal sides (I am not sure this is the standard way for calling them in English, so please edit this answer if needed) are $C$ and $D$ and the parallel sides are $A$ and $B$ with $A>B$, $$ A = \sqrt{C^2-h^2}+\sqrt{D^2-h^2} + B \tag{1}$$ holds by the Pythagorean theorem.
You may just check which permutation of $a,b,c,d$ fulfills this identity.
In $(1)$ I am actually also assuming that the angles on the major base are both acute.
Some signs have to be changed in $(1)$ is this is not the case.
In this case, for instance, we have $$ A = \color{red}{-}\sqrt{C^2-h^2}+\sqrt{D^2-h^2}+B.$$