Inspired by 1989

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3),D(x_4,y_4)\Rightarrow E\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right),F\left(\frac{x_1+x_4}{2},\frac{y_1+y_4}{2}\right)$

Using the shoelace formula, we have $S_{\triangle EDA}+S_{\triangle FBC}=\frac{1}{2}\left(\begin{vmatrix} x_1&x_4&\frac{x_2+x_3}{2}&x_1\\y_1&y_4&\frac{y_2+y_3}{2}&y_1 \end{vmatrix}+\begin{vmatrix} x_2&\frac{x_1+x_4}{2}&x_3&x_2\\y_2&\frac{y_1+y_4}{2}&y_3&y_2 \end{vmatrix}\right)=\frac{x_1y_4-x_4y_1+x_4y_3-x_3y_4+x_3y_2-x_2y_3+x_2y_1-x_1y_2}{2}$

By inspection, this also appears to be the area of the quadrilateral when shoelace is applied. Hence $\theta=0$ and the answer is undefined, which is none of the above.