In the given figure, CM and RN are respectively, the medians of ABC and PQR . If ABC PQR , prove that : (i) AMC PNR…

(i) $\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR}$ [$\frac{1}{2}$ mark]
$\Rightarrow \frac{AC}{PR} = \frac{\frac{1}{2} AB}{\frac{1}{2} PQ} \Rightarrow \frac{AC}{PR} = \frac{AM}{PN}$ [$\frac{1}{2}$ mark]
Also $\angle A = \angle P$ [$\frac{1}{2}$ mark]
$\therefore \Delta AMC \sim \Delta PNR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]
(ii) $\Delta AMC \sim \Delta PNR$ (from part (i))
$\therefore \angle ACM = \angle PRN$ [$\frac{1}{2}$ mark]
Also $\angle ACB = \angle PRQ$ (as $\Delta ABC \sim \Delta PQR$)
$\therefore \angle ACB - \angle ACM = \angle PRQ - \angle PRN$
$\Rightarrow \angle BCM = \angle QRN$ [$\frac{1}{2}$ mark]
(iii) $\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{BC}{QR}$ [$\frac{1}{2}$ mark]
$\Rightarrow \frac{BC}{QR} = \frac{\frac{1}{2} AB}{\frac{1}{2} PQ} \Rightarrow \frac{BC}{QR} = \frac{BM}{QN}$ [$\frac{1}{2}$ mark]
Also $\angle B = \angle Q$ [$\frac{1}{2}$ mark]
$\therefore \Delta BMC \sim \Delta QNR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]