Trapezium from two diagonals and two non-parallel sides Is it possible to construct (and to calculate) a trapezium from it's two non-parallel sides and it's two diagonals, with other words $b,d,e,f$ are given:

I read out the equations system 
$\begin{array}{|l l}
(1) & a =c+p+q \\
(2) & h^2 =e^2 -(a-q)^2 \\
(3) & h^2 =f^2 -(a-p)^2 \\
(4) & b^2 =h^2+q^2 \\
(5) & d^2 =h^2+p^2 \\
\end{array}$ 
How can I solve that for $a,c,h,p,q$ ?
 A: Per the cosine rule
$$a^2 = b^2+e^2 -2be \cos C= d^2 +f^2 -2df \cos D\tag1
$$
and note that the triangles ADB and ACB are equal in area
$$\frac12 be \sin C =\frac12df \sin D\implies b^2e^2 (1-\cos^2 C )= d^2f^2(1-\cos^2 D)\tag2
$$
Substitute (1) into (2) to eliminate $\cos C$ and $\cos D$, yielding
$$a^2= \frac{(b^2-e^2)^2- (d^2-f^2)^2}{2(b^2+e^2 -d^2-f^2)}
$$
Similarly
$$c^2= \frac{(b^2-f^2)^2- (d^2-e^2)^2}{2(b^2+f^2 -d^2-e^2)}
$$
Depending on the given values of $b$, $d$, $e$ and $f$, the trapezium is either unique or impossible since the RHS’s of above expressions have to be positive. In the special case of $b=d$ and $e=f$, the trapezium is possible yet not unique.
A: It takes five dimensions ( angle or side lengths ) to construct a quadrilateral. With four dimensions quadrilaterals can be constructed only upto a single parameter rotation of members. There can be no unique construction.
In kinematics  the situation is referred to as a 4-Bar mechanism configuration.
Constructions for direct and crossed diagonals ( asked for in the question) are given in a rough sketch where the side/link AB is taken fixed (either end does not move) among all possibilities
Thick green two crossed diagonals cases and black direct four diagonal cases are sketched below. Possible configurations are infinitely many.
Makes no difference ... a trapezium or a quadrilateral..  diagonals direct or crossed... the degree of freedom is one.
EDIT1:
Even in the non-trapezium "scalene" quadrilateral of straight diagonals case the connecting rod CD passes through an instant where AB is parallel to CD. This is also true when diagonals are crossed

A: At least in the case when the side lengths are equal and the diagonals are equal, this is impossible.
Attached are three cases with side length $AD = FD' = 4$ and diagonal length $AD' = FD = 5$:



