smc

Because $ABCD$ is a convex quadrilateral, the sum of its interior angles is $360^{\circ }$. Thus, $S^{\prime }=\measuredangle BAD+\measuredangle ABC=360^{\circ }-(\measuredangle BCD+\measuredangle ADC)$. Furthermore, because $\angle BCD$ and $\angle ADC$ are supplementary to $\angle DCE$ and $\angle CDE$, respectively, the four angles sum to $2\ast 180^{\circ }=360^{\circ }$, so $\measuredangle BCD+\measuredangle ADC=360^{\circ }-(\measuredangle DCE+\measuredangle CDE)=360^{\circ }-S$. Plussing this expression for $\measuredangle BCD+\measuredangle ADC$ into the first equation, we see that $S^{\prime }=360^{\circ }-(360^{\circ }-S)=S$, so $r=\frac{S}{S^{\prime }}=1$, which is answer choice $\boxed{r=1}$.