jmc

Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$. Since $AB\parallel CE,BC\parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $△ABC\cong △CFA$. Also, as $AC\parallel DE$, it follows that $△ABC\sim △EFD$. By the Law of Cosines, $A{C}^2=3^2+5^2-2\cdot 3\cdot 5\cos 120^{\circ }=49⟹AC=7$. Thus the length similarity ratio between $△ABC$ and $△EFD$ is $\frac{AC}{ED}=\frac{7}{15}$. Let $h_{ABC}$ and $h_{BDE}$ be the lengths of the altitudes in $△ABC,△BDE$ to $AC,DE$ respectively. Then, the ratio of the areas $\frac{[ABC]}{[BDE]}=\frac{\frac{1}{2}\cdot h_{ABC}\cdot AC}{\frac{1}{2}\cdot h_{BDE}\cdot DE}=\frac{7}{15}\cdot \frac{h_{ABC}}{h_{BDE}}$. However, $h_{BDE}=h_{ABC}+h_{CAF}+h_{EFD}$, with all three heights oriented in the same direction. Since $△ABC\cong △CFA$, it follows that $h_{ABC}=h_{CAF}$, and from the similarity ratio, $h_{EFD}=\frac{15}{7}{h}_{ABC}$. Hence $\frac{h_{ABC}}{h_{BDE}}=\frac{h_{ABC}}{2h_{ABC}+\frac{15}{7}{h}_{ABC}}=\frac{7}{29}$, and the ratio of the areas is $\frac{7}{15}\cdot \frac{7}{29}=\frac{49}{435}$. The answer is $m+n=\boxed{484}$.