Question

CM and RN are respectively the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC\sim \Delta PQR$, prove that:

(i) $\Delta AMC\sim \Delta PNR$

(ii) $\frac{CM}{RN}=\frac{AB}{PQ}$

(iii) $\Delta CMB\sim \Delta RNQ$

Answer 1

$\Delta ABC\sim \Delta PQR$ [Given]

$\Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}......\left(1\right)$

$\angle A=\angle P,\angle B=\angle Qand\angle C=\angle R......\left(2\right)$

(i) In $\Delta AMCand\Delta PNR$

2 AM = AB and 2 PN = PQ [$\because$ CM and RN are medians]

$\Rightarrow \frac{2AM}{2PN}=\frac{CA}{RP}$ [from (1)]

$\Rightarrow \frac{AM}{PN}=\frac{CA}{RP}$

Also, $\angle MAC=\angle NPR$ [From (2)]

$\therefore \Delta AMC\sim \Delta PNR$ [SAS similarity]

(ii) Since $\Delta AMC\sim \Delta PNR$

$\frac{CM}{RN}=\frac{CA}{RP}$

But $\frac{CA}{RP}=\frac{AB}{PQ}$ [From (1)]

$\Rightarrow \frac{CM}{RN}=\frac{AB}{PQ}$

(iii) Again,

$\frac{AB}{PQ}=\frac{BC}{QR}$ [From (1)]

$\therefore \frac{CM}{RN}=\frac{BC}{QR}$

Also,

$\frac{CM}{RN}=\frac{AB}{PQ}=\frac{2BM}{2QN}$

$\Rightarrow \frac{CM}{RN}=\frac{BM}{QN}$

$\frac{CM}{RN}=\frac{BC}{QR}=\frac{BM}{QN}$

$\therefore \Delta CMB\sim \Delta RNQ$ [SSS similarity]

[Note : You can also prove part (iii) by following the same method as used for proving part (i)]