How would one set up barycentric coordinates for a trapezoid? Barycentric coordinates are great for triangles, but I'm interested in how to construct a barycentric coordinate system for an arbitrary trapezoid. 
I've seen this done for an arbitrary quadrilateral, but it ought to be simpler in the case of a trapezoid because of the two parallel sides. There does not appear to be information on this specific case online. 
This source has a good solution in general, but has this note: "Note that the special case $\vec{c}\times\vec{d}=0$ must be treated separately", but gives no explanation of how the special case must be treated. This special case turns out to be when two opposite sides are parallel — the definition of a trapezoid. 

The source derives that
$$(\mathbf{c} \times \mathbf{d})\mu^2 + (\mathbf{c} \times \mathbf{b} + \mathbf{a} \times \mathbf{d})\mu + \mathbf{a} \times \mathbf{b} = 0.$$
When $\mathbf{c} \times \mathbf{d}$ is zero, we should just be able to conclude that
$$\mu = \frac{\mathbf{a} \times \mathbf{b}}{\mathbf{c} \times \mathbf{b} + \mathbf{a} \times \mathbf{d}}.$$
I think perhaps they mean that one cannot apply the quadratic formula in this case.
