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AD and PS are respectively, the medians of $\triangle ABC$ and $\triangle PQR$. If $\triangle ABC \sim \triangle PQR$, then prove that
(i) $\triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{BC}{QR}$
(i) $\triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{BC}{QR}$
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OR
(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} = \frac{2DC}{2SR}$ (2 Marks)
$\Rightarrow \frac{AC}{PR} = \frac{DC}{SR}$
and $\angle C = \angle R$
$\therefore \triangle ADC \sim \triangle PSR$ (by SAS similarity) (1/2 Mark)
(ii) Since $\triangle ADC \sim \triangle PSR$, $\frac{AD}{PS} = \frac{DC}{SR}$ (1 Mark)
$\frac{AD}{PS} = \frac{BC/2}{QR/2} = \frac{BC}{QR}$ (1 Mark)
(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} = \frac{2DC}{2SR}$ (2 Marks)
$\Rightarrow \frac{AC}{PR} = \frac{DC}{SR}$
and $\angle C = \angle R$
$\therefore \triangle ADC \sim \triangle PSR$ (by SAS similarity) (1/2 Mark)
(ii) Since $\triangle ADC \sim \triangle PSR$, $\frac{AD}{PS} = \frac{DC}{SR}$ (1 Mark)
$\frac{AD}{PS} = \frac{BC/2}{QR/2} = \frac{BC}{QR}$ (1 Mark)