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?

In the case $a = \pi/2$, $x = \arccos \left( 2\,{\frac {\sin \left( b \right) }{\sqrt {4-3\, 
\cos^2 \left( b \right)   }}} \right)$.  This is not an algebraic function of $b$, because its derivative is $\frac{dx}{db} = \frac{2}{3 \cos^2 b - 4}$ for $-\pi/2 < b < \pi/2$, and $\cos(b)$ is not an algebraic function. 
