Broken Pinwheel

The sides AB and CD of the quadrilateral have to be parallel, since $\angle ABD=\angle BDC$ . We can set the distance between the lines to 1. Let's name the $\angle BAC=\angle ACD=x$ . This has to satisfy the equation

$\cot(x+40^\circ)+1+\cot(x+20^\circ)=\cot(x)$

Using the identity for a cotangent of a sum, we get

$\cot(x)-1=\frac{\cot(40^\circ)\times \cot(x)-1}{\cot(40^\circ)+\cot(x)}+\frac{\cot(20^\circ)\times \cot(x)-1}{\cot(20^\circ)+\cot(x)}$

This is a third degree equation in $\cot(x)$ with three solutions:

$\cot(x_{1})=2.7858, x_{1}=19.7463^\circ$ , which corresponds to an angle between diagonals of $64.7463^\circ$ .

$\cot(x_{2})=0.1249, x_{2}=82.8797^\circ$ , which corresponds to an angle between diagonals of $52.1203^\circ$ .

and finally $\cot(x_{3})=-1.9071, x_{3}=117.6701^\circ$ .

This configuration, however, does not give a quadrilateral with desired properties and is therefore not a valid solution.