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(i) If $\triangle ABC \sim \triangle PQR$, then prove that $\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle PQR)} = (\frac{AB}{PQ})^2 = (\frac{BC}{QR})^2 = (\frac{CA}{RP})^2$.
(ii) In the given figure, $\angle ADC = \angle BAC$. If AC = 8 cm and AD = 3 cm, then find the length of AB.
(ii) In the given figure, $\angle ADC = \angle BAC$. If AC = 8 cm and AD = 3 cm, then find the length of AB.
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(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$ (2 Marks)
(ii) Since $\triangle ADC \sim \triangle BAC$ (by SAS similarity) (1 Mark)
$\frac{AC}{BC} = \frac{DC}{AC} = \frac{AD}{BA}$ (1 Mark)
$\frac{8}{BC} = \frac{DC}{8} = \frac{3}{BA}$ (1 Mark)
(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$ (2 Marks)
(ii) Since $\triangle ADC \sim \triangle BAC$ (by SAS similarity) (1 Mark)
$\frac{AC}{BC} = \frac{DC}{AC} = \frac{AD}{BA}$ (1 Mark)
$\frac{8}{BC} = \frac{DC}{8} = \frac{3}{BA}$ (1 Mark)