In a $\triangle ABC$, D is a point on AC such that AD=2 cm, DC = 6 cm and $\angle ADB = \angle ABC$. Which of the following cannot be the length of BC?

D is a point on AC, so:

$AC = AD + DC = 2 + 6 = 8$ cm

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In $\triangle ADB$ and $\triangle ABC$:

$\angle A = \angle A$ (common angle)

$\angle ADB = \angle ABC$ (given)

By AA Similarity, $\triangle ADB \sim \triangle ABC$

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Since the triangles are similar, corresponding sides are proportional:

$\dfrac{AD}{AB} = \dfrac{AB}{AC}$

$AB^2 = AD \times AC$

$AB^2 = 2 \times 8 = 16$

$AB = 4$ cm

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Now in $\triangle ABC$, $AB = 4$ cm and $AC = 8$ cm.

By the triangle inequality, the third side must satisfy:

$|AC - AB| < BC < AC + AB$

$|8 - 4| < BC < 8 + 4$

$4 < BC < 12$

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$BC = 11$ cm lies in $(4, 12)$ — possible

$BC = 10$ cm lies in $(4, 12)$ — possible

$BC = 12$ cm does not lie in $(4, 12)$ since $BC$ must be strictly less than $12$ — not possible

$BC = 9$ cm lies in $(4, 12)$ — possible

Therefore, $BC = 12$ cm cannot be the length of $BC$.