Computing the lengths of the obtained trapezium $ABCD$ is a quadrilateral. A line through $D$ parallel to $AC$ meets $BC$ produced at $P$. My book asked me to show that the area of $APB$ and $ABCD$, are the same, which I did. But it aroused my interest. So I researched on how do we compute the sides or simply characterize the trapezium $ACPD$, if we know the sides and angles of $ABCD$. But I was not able to find any result. Please help.
 A: Edit：

It is easy find all for $ACPD$, $AC$ is trivial to get. then you can have $F$, since $AB$ is known,so $AF,FB,FC$ can be obtained. then you can find $CP,DP$ since $AC$  \\  $DP$
A: Refer to the following figure:-

Notation:-  [Object] = area of that object.
using the formula, ⊿ = (1/2) ab sin θ, 
[quad ABCD] = … 
= f(a, b, c, d, θ, Ø)
= f, a function depending on the mentioned variables
[ABP] = (1/2) H (b + x) = (1/2) a sin θ (b + x)
[quad ABCD] = [ABP] implies  $\frac{a \sin \theta (b + x)}{ 2} = f$
∴ $(b + x) = \frac {2f} {\sin \theta}$ or $x = \frac {2f} {\sin \theta} – b$
Thus, ⊿ABP can be completely solved, provided a, b, c, d, θ, Ø are known.
We now try to see if some of the givens can be relaxed.
(1) If ABCD is a trapezium with AD // BC, the requirement can be relaxed to a, b, c, d, θ only because [ABCD] can be calculated by the extended Heron formula without the presence of Ø.
(2) In fact, As AC = g(a, b, θ), using the cosine law.
And Ø = h(AC, c, d).
This means Ø need not be known.
Hence, we can conclude that:- 
If a quadrilateral has its 4 sides and an angle known, the components of the transformed triangle having the same area can be completely soklved.
