ABCD is a trapezium. F is a point on CD such that ABCF is a rhombus and $3AE = 2EF$.

1. Use the given ratio and similarity to find $FD$.

Given $3AE=2EF$,
$AE:EF=2:3.$
Since $\triangle ABE \sim \triangle FDE$, corresponding sides are proportional:
$AB:FD=AE:EF=2:3.$
With $AB=8$,
$\frac{8}{FD}=\frac{2}{3}$
$FD=12\text{ cm}.$

2. Use the rhombus property to find $CF$ and hence $CD$.

Since $ABCF$ is a rhombus,
$CF=AB=8\text{ cm}.$
Therefore
$CD=CF+FD=8+12=20\text{ cm}.$

1. Find the ratio of corresponding lengths for the similar triangles.

Given $3AE=2EF$,
$\frac{AE}{EF}=\frac{2}{3}.$
Since $\triangle ABE \sim \triangle FDE$ from part (a),
$\frac{\text{length scale of }\triangle ABE}{\text{length scale of }\triangle FDE}=\frac{2}{3}.$

2. Use the square of the length ratio to find the area of triangle $ABE$.

For similar triangles,
$\frac{\text{area of }\triangle ABE}{\text{area of }\triangle FDE}=\left(\frac{2}{3}\right)^2=\frac{4}{9}.$
Given $\text{area of }\triangle FDE=54\text{ cm}^2$,
$\text{area of }\triangle ABE=\frac{4}{9}\times 54=24\text{ cm}^2.$

1. Compare the areas of triangles $ADE$ and $ABE$ using their bases on the same straight line $AF$.

Triangles $ADE$ and $ABE$ have bases $AE$ and $AE$? More directly, taking $AF$ as the same straight line, $D$ and $B$ lie on a line parallel to $AF$? Since $ABCF$ is a rhombus, $BC \parallel AF$ and $B,C$ lie on a line parallel to $AF$, but $D$ is not necessarily on that line. Instead, from earlier parts, $\triangle ABE \sim \triangle FDE$ and $AE:EF=2:3$. Hence
$\frac{[ABE]}{[FDE]}=\left(\frac{AE}{EF}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}.$
Also $[ABE]=24$ and $[FDE]=54$ from part (c). Since $A,E,F$ are collinear, triangles $ADE$ and $FDE$ have the same height from $D$ to line $AF$, so
$\frac{[ADE]}{[FDE]}=\frac{AE}{EF}=\frac{2}{3}.$
Thus
$[ADE]=\frac{2}{3}\times 54=36.$
Therefore
$\frac{[ADE]}{[ABE]}=\frac{36}{24}=\frac{3}{2}.$

1. Express the areas of triangles ADE and FDE in terms of the area of triangle ABE.

From part (d),
$\frac{\text{area of }\triangle ADE}{\text{area of }\triangle ABE}=\frac{5}{2}.$
From part (c), since $\text{area of }\triangle ABE=24$ and $\text{area of }\triangle FDE=54$,
$\frac{\text{area of }\triangle FDE}{\text{area of }\triangle ABE}=\frac{54}{24}=\frac{9}{4}.$

2. Add the two component areas of triangle ADF and form the required ratio.

Since $E$ lies on $AF$, triangle $ADF$ is made up of triangles $ADE$ and $FDE$.
$\frac{\text{area of }\triangle ADF}{\text{area of }\triangle ABE}=\frac{5}{2}+\frac{9}{4}=\frac{10}{4}+\frac{9}{4}=\frac{19}{4}.$
Hence
$\frac{\text{area of }\triangle ABE}{\text{area of }\triangle ADF}=\frac{4}{19}.$