(A) In the given figure, ∠ ADC =∠ BCA; prove that Δ ACB ∼ Δ ADC. Hence find BD if AC =8 cm and AD =3 cm.

In \(\displaystyle \Delta \mathrm{ACB}\) and \(\displaystyle \Delta \mathrm{ADC}\),

\[\begin{aligned} & \angle \mathrm{ACB}=\angle \mathrm{ADC} \\ & \angle \mathrm{~A}=\angle \mathrm{A} \\ & \therefore \Delta \mathrm{ACB} \sim \Delta \mathrm{ADC} \\ & \therefore \frac{\mathrm{AC}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{AC}} \Rightarrow \frac{8}{3}=\frac{\mathrm{AB}}{8} \\ & \Rightarrow \mathrm{AB}=\frac{64}{3} \\ & \mathrm{AB}-\mathrm{AD}=\frac{64}{3}-3=\frac{55}{3} \mathrm{~cm} \end{aligned} \]

Correct given, to prove, figure and construction

Correct Proof