If $x$ and $y$ are adjacent, without loss of generality, we need to show that $S \leqslant \frac{1}{2}(a b+c d)$. This follows from $S_{A B C}=\frac{1}{2} A B \cdot B C \cdot \sin \widehat{A B C} \leqslant \frac{1}{2} a b$, and similarly $S_{C D A} \leqslant \frac{1}{2} c d$. If $x$ and $y$ are opposite sides, we need to show that $S \leqslant \frac{1}{2}(a c+b d)$. Let $A^{\prime}$ be the symmetric point of $A$ with respect to the perpendicular bisector of $[B D]$. Then $B A^{\prime} D$ is isometric to $D A B$. We apply the above to $A^{\prime} B C D$, which gives $S=S_{B A D}+S_{B C D}=S_{B A^{\prime} D}+S_{B C D}=S_{A^{\prime} B C D} \leqslant \frac{1}{2}\left(A^{\prime} B \cdot B C+\right.$ $\left.C D \cdot D A^{\prime}\right)=\frac{1}{2}(a c+b d)$.