A geometry problem (length of side of a quadrilateral) This problem showed up on my tab and I have tried few approaches, but they don't span out. So this is the figure,
In this figure we have find the distance BC, in terms of AD, AB and $\theta$ only.
Any pointers would be helpful.
P.S: In one of my methods I drew diagonals and used cosine rule, then cross-substituted the equations.
 A: HINT:

*

*Join BD.

*$\angle BDA=\angle CBD-\theta$ (that’s the best I could explain which angle I’m talking about) as Alternate Angles wrt a transversal (here, BD) are equal.

*$\angle BDA=\angle CBD-\theta=\tan^{-1}\left(\dfrac{BA}{AD}\right)$.

*$\triangle CBD$ is a right triangle.

*$\cos\angle CBD=\cos\left(\tan^{-1}\left(\dfrac{BA}{AD}\right)+\theta\right)=\dfrac{BC}{BD}=\dfrac{BC}{\sqrt{AB^2+AD^2}}$.

*$BC=\sqrt{AB^2+AD^2}\cdot\cos\left(\tan^{-1}\left(\dfrac{BA}{AD}\right)+\theta\right)$.

A: Drop a perpendicular from $A$ to $CD.$ Call the intersection point $E.$
Drop a perpendicular from $B$ to $AE.$ Call the intersection point $F.$
You now have partitioned the original figure into two right triangles and a rectangle, as shown below:

(The rectangle doesn't look very rectangular, but that's mainly because of the not-to-scale right angle at $C.$)
You should now be able to find all the lengths you need as quite simple expressions involving $AB,$ $AD,$ and trigonometric functions of $\theta.$
