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The problem seems to describe a triangle \( ABC \) with point \( D \) located on side \( BC \), where the segment divides the triangle into two smaller triangles \( \triangle ABD \) and \( \triangle ADC \). From the text, we know:

1. \( BD = 4 \, \text{cm} \)
2. \( DC = 8 \, \text{cm} \)
3. The ratio of the areas is given: \( \frac{S_{ADC}}{S_{ABD}} = 2 \)

Let's analyze and solve this.

### Area Ratio:
The areas of triangles \( \triangle ABD \) and \( \triangle ADC \) depend on both their bases and their heights. Since \( \triangle ABD \) and \( \triangle ADC \) share the same height from point \( A \) to side \( BC \), the ratio of their areas will be proportional to the lengths of their bases. Specifically:
\[
\frac{S_{ADC}}{S_{ABD}} = \frac{DC}{BD} = 2
\]
From the given values, \( BD = 4 \, \text{cm} \) and \( DC = 8 \, \text{cm} \), we can confirm that:
\[
\frac{DC}{BD} = \frac{8}{4} = 2
\]
This matches the given area ratio \( \frac{S_{ADC}}{S_{ABD}} = 2 \), which verifies the relationship.

### Conclusion:
The division of triangle \( ABC \) by point \( D \) on side \( BC \) leads to two triangles, where the area of \( \triangle ADC \) is twice that of \( \triangle ABD \), and the ratio of the segments \( BD \) to \( DC \) reflects this.

Would you like further clarification on this, or do you have any other questions? Here are 5 related questions to deepen understanding:

1. How would the area ratio change if the lengths of \( BD \) and \( DC \) were different?
2. Can you calculate the total length of \( BC \)?
3. What is the significance of the height from \( A \) to \( BC \) in determining the area ratio?
4. How would you calculate the actual areas of \( \triangle ABD \) and \( \triangle ADC \)?
5. What would happen to the area ratio if \( D \) were placed at the midpoint of \( BC \)?

**Tip:** When dividing a triangle into two smaller triangles by a point on one side, the area ratio between the two triangles is directly proportional to the ratio of the divided side segments.

Given a triangle ABC, point D is located on side BC. The segment divides triangle ABC into two smaller triangles ABD and ADC. The following values are provided: BD = 4 cm, DC = 8 cm. The ratio of the areas of triangles ADC and ABD is 2. Prove the given ratio.

This problem involves proving the ratio of the areas of two triangles within a larger triangle divided by a point on one side. Using the given side lengths BD = 4 cm and DC = 8 cm, the solution demonstrates how to establish that the area of triangle ADC is twice that of triangle ABD.

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