The diagram shows a triangle AGD with the points C and B lying on AD and points E and F lying on GD. It is given that $CE//BF//AG$, $BF = 2CE$ and $AG = 3CE$.

1. Identify one pair of equal angles at the common vertex $D$.

Since $C$ and $B$ lie on the same straight line $AD$, and $E$ and $F$ lie on the same straight line $GD$,
$\angle CDE = \angle BDF.$

2. Identify a second pair of equal corresponding angles using the parallel lines.

Since $CE \parallel BF$ and $DC$ is on the same straight line as $DB$,
$\angle DCE = \angle DBF$
corresponding angles.

3. Conclude similarity by AA.

Therefore, two pairs of corresponding angles are equal. Hence,
$\triangle DCE \sim \triangle DBF \quad \text{(AA similarity).}$

1. Use the parallel-line length ratios to find the linear scale factors of the similar triangles.

From part (a), $\triangle DCE \sim \triangle DBF$.

Since $BF=2CE$, the linear scale factor from $\triangle DCE$ to $\triangle DBF$ is $2$.

Also, $AG=3CE$, so $\triangle DCE \sim \triangle DAG$ with linear scale factor $3$.

2. Convert linear scale factors to area ratios.

Areas of similar figures are in the square of their corresponding side ratios.

$[DCE]:[DBF]:[DAG]=1^2:2^2:3^2=1:4:9$

3. Use the given trapezium area to find the area of the larger trapezium.

Trapezium $BFEC$ is the part between $\triangle DBF$ and $\triangle DCE$.

$[BFEC]=[DBF]-[DCE]=4-1=3\text{ parts}$

Given $[BFEC]=15\text{ cm}^2$,

$1\text{ part}=\frac{15}{3}=5\text{ cm}^2$

Trapezium $AGFB$ is the part between $\triangle DAG$ and $\triangle DBF$.

$[AGFB]=[DAG]-[DBF]=9-4=5\text{ parts}$

$[AGFB]=5\times 5=25\text{ cm}^2$