Computing the lengths of the obtained trapezium

Question

$ABCD$ 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.

Answer 1

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.

Answer 2

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 $AC$ \\ $DP$