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In the given figure, CD and RS 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) $AD \times PR = AC \times PS$
(i) $\triangle ADC\sim\triangle PSR$
(ii) $AD \times PR = AC \times PS$
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(i) $\triangle ABC \sim \triangle PQR$
$\angle A=\angle P$
and $\frac{AB}{PQ} = \frac{AC}{PR}$
$\Rightarrow \frac{2AD}{2PS} = \frac{AC}{PR}$
$\Rightarrow \frac{AD}{PS} = \frac{AC}{PR}$ and $\angle A = \angle P$
Therefore $\triangle ADC \sim \triangle PSR$
(ii)Hence $\frac{AD}{PS} = \frac{AC}{PR}$
$\Rightarrow AD \times PR = AC \times PS$
$\angle A=\angle P$
and $\frac{AB}{PQ} = \frac{AC}{PR}$
$\Rightarrow \frac{2AD}{2PS} = \frac{AC}{PR}$
$\Rightarrow \frac{AD}{PS} = \frac{AC}{PR}$ and $\angle A = \angle P$
Therefore $\triangle ADC \sim \triangle PSR$
(ii)Hence $\frac{AD}{PS} = \frac{AC}{PR}$
$\Rightarrow AD \times PR = AC \times PS$