Given: CM and RN are medians of △ABC and △PQR respectively. △ABC ~ △PQR.
To Prove: (i) △AMC ~ △PNR (ii) △CMB ~ △RNQ
---
Proof of (i): △AMC ~ △PNR
Since △ABC ~ △PQR:
$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \quad \text{...(1)}$$
$$\angle A = \angle P,\ \angle B = \angle Q,\ \angle C = \angle R \quad \text{...(2)}$$
Since CM and RN are medians, M and N are midpoints of AB and PQ respectively.
$$\therefore AB = 2AM \text{ and } PQ = 2PN$$
From (1): $\dfrac{2AM}{2PN} = \dfrac{CA}{RP}$
$$\Rightarrow \frac{AM}{PN} = \frac{CA}{RP} \quad \text{...(3)}$$
Also, $\angle MAC = \angle NPR$ [i.e., $\angle A = \angle P$, from (2)] ...(4)
From (3) and (4), by SAS similarity criterion:
$$\boxed{\triangle AMC \sim \triangle PNR}$$
---
Proof of (ii): △CMB ~ △RNQ
Since M and N are midpoints: $AB = 2BM$ and $PQ = 2QN$
From (1): $\dfrac{2BM}{2QN} = \dfrac{BC}{QR}$
$$\Rightarrow \frac{BM}{QN} = \frac{BC}{QR} \quad \text{...(5)}$$
From part (i), △AMC ~ △PNR, so:
$$\frac{CM}{RN} = \frac{CA}{RP} = \frac{AB}{PQ} = \frac{BC}{QR} \quad \text{...(6)}$$
From (5) and (6):
$$\frac{BC}{QR} = \frac{BM}{QN} = \frac{CM}{RN}$$
By SSS similarity criterion:
$$\boxed{\triangle CMB \sim \triangle RNQ}$$
Source: Chapter 6, Section 6.4 (Example 8)
---