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In the given figure, $\angle ADC = \angle BCA$; prove that $\triangle ACB \sim \triangle ADC$. Hence find BD if $AC = 8$ cm and $AD = 3$ cm.
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In $\triangle ACB$ and $\triangle ADC$,
$\angle ACB = \angle ADC$
$\angle A = \angle A$
$\therefore \triangle ACB \sim \triangle ADC$
$\therefore \frac{AC}{AD} = \frac{AB}{AC} \Rightarrow \frac{8}{3} = \frac{AB}{8}$
$\Rightarrow AB = \frac{64}{3}$
$BD = AB – AD = \frac{64}{3} - 3 = \frac{55}{3}$ cm
$\text{OR}$._a_1.png)
$\angle ACB = \angle ADC$
$\angle A = \angle A$
$\therefore \triangle ACB \sim \triangle ADC$
$\therefore \frac{AC}{AD} = \frac{AB}{AC} \Rightarrow \frac{8}{3} = \frac{AB}{8}$
$\Rightarrow AB = \frac{64}{3}$
$BD = AB – AD = \frac{64}{3} - 3 = \frac{55}{3}$ cm
$\text{OR}$