We are given a trapezium $ABCD$ with $AD \parallel BC$ and $AD = 4 \text{ cm}$. The diagonals $AC$ and $BD$ intersect at point $O$, such that $\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}$, and we are asked to find the length of side $BC$.

Step-by-step approach:

1. Since $AD \parallel BC$, triangles $AOD$ and $BOC$ formed by the diagonals are similar by the AA criterion of similarity (they share an angle at $O$, and corresponding angles are equal due to the parallelism).

2. The ratio $\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}$ suggests that point $O$ divides both diagonals in the same ratio. This means that the triangles $\triangle AOD$ and $\triangle BOC$ are similar, with a scale factor of $1:2$.

3. In similar triangles, the corresponding sides are proportional. Therefore, the ratio between the lengths of $AD$ and $BC$ is also $1:2$.

4. Since $AD = 4 \text{ cm}$, it follows that: $BC = 2 \times AD = 2 \times 4 = 8 \text{ cm}.$

Thus, the length of $BC$ is $\boxed{8} \text{ cm}$.

Would you like more details on this solution or have any further questions?

Related Questions:

1. What are the properties of trapeziums and how do they apply to this problem?
2. How can the concept of similar triangles be used to solve geometric problems?
3. What does it mean for diagonals to divide each other in a given ratio?
4. How does the AA similarity criterion work in identifying similar triangles?
5. How can parallelism in geometric shapes help solve length-related problems?

Tip:

When solving problems involving similar triangles, always look for corresponding sides and angles, and remember that the ratio of the sides will be constant across the triangles.