You Need To Look Everywhere

Draw a line DF, which parallels to AB and EC. Mark all the information provided, you will have a figure like the following:

Note that the length of BF and FC are in ratio 8:7 (Intercept theorem, DF//EC). △BCD and △ECD share the same height DC, so that the ratio of the area of △BCD and the area of △ECD is also 8:7.

Note: The following areas are in ratios only.

△BCD is divided into two triangles by DF, △BDF and △FDC.

The area of △BDF = $\frac{8･8}{8+7}$ = $\frac{64}{15}$

The area of △FDC = $\frac{8･7}{8+7}$ = $\frac{56}{15}$

As ABFD is a parallelogram, area of △BDF = △BDA = $\frac{64}{15}$

Area of △ABD : △CDE = $\frac{64/15}{7}$ = 64 : 105

Let $[ABD]=x \ , \ [CDE]=y$ so the problem asks us to find $\dfrac{x}{y}$ .

Since $AD \ || \ BC$ we have $[ABD]=[ADC]=x$ as they have same base and same height.

Since $\dfrac{BD}{DE}=\dfrac{8}{7}$ we have two area relations:

$\begin{cases} \dfrac{[ABD]}{[ADE]}=\dfrac{8}{7} \implies \dfrac{x}{[ADE]}=\dfrac{8}{7} \implies [ADE]=\dfrac{7x}{8} \\ \dfrac{[BDC]}{[CDE]}=\dfrac{8}{7} \implies \dfrac{[BDC]}{y}=\dfrac{8}{7} \implies [BDC]=\dfrac{8y}{7} \end{cases}$

Now since $AB \ || \ CE$ we have:

$[BCE]=[ACE]$

$\implies [BDC]+[CDE]=[ADC]+[ADE]+[CDE]$

$\implies [BDC]=[ADC]+[ADE]$ Substituting the above computed areas in terms of $x,y$ :

$\implies \dfrac{8y}{7}=x+\dfrac{7x}{8} \implies \dfrac{8y}{7} = \dfrac{15x}{8}$

$\implies \dfrac{x}{y} = \dfrac{8\times 8}{15\times 7} = \boxed{\dfrac{64}{105}}$ which was what we had to find. Cheers!

Notation: Denote $[ABC]$ as the area of $\Delta ABC$ .

Given two side-lengths $x$ and $y$ and the contained angle $\theta$ , the area of a triangle is $\frac{1}{2}xy\sin(\theta)$ .

Since $\angle ABD = \angle DEC$ and $\overline{AB} = \frac{8}{15}\overline{CE}$ , the ratio of the areas is: $\frac{\frac{1}{2}\cdot 8\cdot\frac{8}{15}\overline{CE}\cdot\sin(\angle ABD)}{\frac{1}{2}\cdot 7\cdot\overline{CE}\cdot\sin(\angle ABD)} = \boxed{\frac{64}{105}}$

By the way, the term "trapezium" is defined differently in the U.S. and the U.K.