1977 AMC 12 Problem 26

Split the quadrilateral along each diagonal and bound every sine by $1$

Combine the two resulting area bounds and analyze when every bound is an equality

Splitting along diagonal $MP$ and using $\sin\theta\le1$ gives $A\le\frac{ab+cd}{2}.$ Splitting along $NQ$ similarly gives $A\le\frac{ad+bc}{2}.$ These bounds remain valid for a nonconvex quadrilateral by taking the appropriate difference of triangle areas. Adding them yields $$\begin{aligned} 2A&\le\frac{ab+ad+bc+cd}{2}\\ &=\frac{(a+c)(b+d)}2, \end{aligned}$$ or $A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right).$ Equality requires equality in all four sine bounds, so all four angles are right angles; conversely a rectangle gives equality. Thus the equality holds exactly for rectangles.

Therefore, the correct answer is B.