Finding angles in a parallelogram without trigonometry  
I'm wondering whether it's possible to solve for $x^{\circ}$ in terms of $a^{\circ}$ and $b^{\circ}$ given that $ABCD$ is a parallelogram. In particular, I'm wondering if it's possible to solve it using only "elementary geometry". I'm not sure what "elementary geometry" would usually imply, but I'm trying to solve this problem without trigonometry.
Is it possible? Or if it's not, is there a way to show that it isn't solvable with only "elementary geometry" techniques?

On taking into account the triangles ACP and ABP, where P is the point where the diagonals intercept, we have, from sines' law:  $\sin{x}/\sin{a} = \sin{\theta}/\sin{b}$, $\theta$ being the angle between the smaller diagonal and the line AB. 
Then $\sin{\theta} = \sin{x} \, \dfrac{\sin{b}}{\sin{a}}$. 
Now, note that, in triangle ACP, the sum of the inner angles is $\,x+a+b+\theta = 180^{\circ}$, hence $\sin{\theta} = \sin{(x+a+b)}$. 
Therefore: $\sin{x} \, \dfrac{\sin{b}}{\sin{a}} = \sin{(x+a+b)} = \sin{x} \, \cos{(a+b)} + \sin{(a+b)} \, \cos{x}$. 
This implies that $\cot{x} = \dfrac{\sin{b}}{\sin{a} \, \sin{(a+b)}} - \cot{(a+b)}$.
