In Figure,$CM$ and $RN$ are respectively the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC \sim \Delta PQR$,prove that:
$(i)$ $\Delta AMC \sim \Delta PNR$
$(ii)$ $\frac{CM}{RN} = \frac{AB}{PQ}$
$(iii)$ $\Delta CMB \sim \Delta RNQ$

(A) Given: $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,their corresponding sides are proportional and corresponding angles are equal:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \quad ...(1)$
$\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \quad ...(2)$
Since $CM$ and $RN$ are medians,$M$ is the midpoint of $AB$ and $N$ is the midpoint of $PQ$. Therefore,$AB = 2AM$ and $PQ = 2PN$.
$(i)$ From $(1)$,$\frac{2AM}{2PN} = \frac{CA}{RP} \implies \frac{AM}{PN} = \frac{CA}{RP}$.
Also,$\angle MAC = \angle NPR$ (from $(2)$).
By $SAS$ similarity criterion,$\Delta AMC \sim \Delta PNR$.
$(ii)$ Since $\Delta AMC \sim \Delta PNR$,their corresponding sides are proportional:
$\frac{CM}{RN} = \frac{CA}{RP}$.
From $(1)$,$\frac{CA}{RP} = \frac{AB}{PQ}$.
Therefore,$\frac{CM}{RN} = \frac{AB}{PQ}$.
$(iii)$ In $\Delta CMB$ and $\Delta RNQ$:
$\frac{CM}{RN} = \frac{BC}{QR}$ (from $(ii)$ and $(1)$).
$\frac{BC}{QR} = \frac{BM}{QN}$ (since $BC = 2BM$ and $QR = 2QN$ from the median property).
Thus,$\frac{CM}{RN} = \frac{BC}{QR} = \frac{BM}{QN}$.
By $SSS$ similarity criterion,$\Delta CMB \sim \Delta RNQ$.