In the given figure \(\Delta ABC\) is shown, in which \(DE \parallel BC\). If \(AD = 5\text{ cm}\), \(DB = 2\cdot5\text{ cm}\) and \(DE = 8\text{ cm}\), then the length of \(BC\) is :
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Do not confuse Thales' Theorem (Basic Proportionality Theorem) with Triangle Similarity.
BPT states \(\frac{AD}{DB} = \frac{AE}{EC}\), which involves only segments on the sides.
When calculating lengths of parallel parallel bases like \(DE\) and \(BC\), you must use similar triangles: \(\frac{AD}{AB} = \frac{DE}{BC}\).
Step 1: Establish similarity.
Since $DE\parallel BC$, corresponding angles give $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$, so by AA similarity, $\triangle ADE\sim\triangle ABC$.
Step 2: Build the scale factor from AD and DB directly, without first adding them.
The scale factor is $k=\dfrac{AB}{AD}$. Since $AB=AD+DB$:
\[ k = \frac{AD+DB}{AD} = 1+\frac{DB}{AD} \]
Substituting $AD=5$ cm and $DB=2.5$ cm:
\[ k = 1+\frac{2.5}{5} = 1.5 \]
Step 3: Apply the same scale factor to DE to get BC.
Since $\dfrac{BC}{DE}=k$ as well:
\[ BC = k\times DE = 1.5\times8 = 12\text{ cm} \]
Final Answer:
$BC=12$ cm, confirming option (C).
\[ \boxed{12\text{ cm}} \]
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