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In the given figure, $CM$ and $RN$ are respectively the medians of $\triangle ABC$ and $\triangle PQR$.
If $\triangle ABC \sim \triangle PQR$, then prove that :
(i) $\triangle AMC \sim \triangle PNR$
(ii) $\triangle CMB \sim \triangle RNQ$
If $\triangle ABC \sim \triangle PQR$, then prove that :
(i) $\triangle AMC \sim \triangle PNR$
(ii) $\triangle CMB \sim \triangle RNQ$
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(i) $\triangle ABC \sim \triangle PQR \implies \angle A = \angle P$ (1/2 Mark)
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{2 AM}{2 PN} = \frac{AC}{PR}$ (as $CM$ and $RN$ are the medians) (1.5 Marks)
$\therefore \triangle AMC \sim \triangle PNR$ (1/2 Mark)
(ii) $\triangle ABC \sim \triangle PQR \implies \angle B = \angle Q$ (1/2 Mark)
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2 MB}{2 NQ} = \frac{BC}{QR}$ (as $CM$ and $RN$ are the medians) (1.5 Marks)
$\therefore \triangle CMB \sim \triangle RNQ$ (1/2 Mark)
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{2 AM}{2 PN} = \frac{AC}{PR}$ (as $CM$ and $RN$ are the medians) (1.5 Marks)
$\therefore \triangle AMC \sim \triangle PNR$ (1/2 Mark)
(ii) $\triangle ABC \sim \triangle PQR \implies \angle B = \angle Q$ (1/2 Mark)
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2 MB}{2 NQ} = \frac{BC}{QR}$ (as $CM$ and $RN$ are the medians) (1.5 Marks)
$\therefore \triangle CMB \sim \triangle RNQ$ (1/2 Mark)