To solve this, we will analyze the problem using the basic proportionality theorem (BPT), also known as Thales' theorem.

Observations from the diagram:

1. $\triangle ABC$ is given, and $DE \parallel BC$.
2. By the basic proportionality theorem:
   • A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

Verification of each option:

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(a) $\frac{AD}{DB} = \frac{AE}{EC}$:
This is a direct result of the basic proportionality theorem. Since $DE \parallel BC$, the ratios of the divided segments on $AB$ and $AC$ are equal.

• This is correct.

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(b) $\frac{AD}{DB} = \frac{DE}{BC}$:
The left side represents the ratio of segments on $AB$, while the right side compares a segment on the parallel line $DE$ to the base $BC$.

• This is incorrect because the basic proportionality theorem does not relate the segments of the parallel line to the sides of the triangle in this manner.

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(c) $\frac{AD}{AB} = \frac{AE}{AC}$:
This relation follows from the division of the sides of the triangle. Since $AD + DB = AB$ and $AE + EC = AC$, the proportionality holds.

• This is correct.

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(d) $\frac{BD}{BA} = \frac{CE}{CA}$:
Rewriting this as $\frac{DB}{AB} = \frac{EC}{AC}$, this is another valid proportionality relationship derived from the basic proportionality theorem.

• This is correct.

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Conclusion:

The incorrect statement is (b): $\frac{AD}{DB} = \frac{DE}{BC}$.

Let me know if you want a further breakdown or proof for any of the relationships!

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Related Questions:

1. What are the applications of the basic proportionality theorem in real-life problems?
2. How can you find the length of unknown segments using proportionality?
3. What happens when $DE$ is not parallel to $BC$?
4. Can similar triangles be used to verify these relationships?
5. How does the converse of the basic proportionality theorem work?

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Tip:

Whenever you see parallel lines in triangles, always think of the basic proportionality theorem for quick ratio relationships!