If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR,$ respectively,where $\Delta ABC \sim \Delta PQR,$ prove that $\frac{AB}{PQ} = \frac{AD}{PM}.$

(N/A) It is given that $\Delta ABC \sim \Delta PQR.$
We know that the corresponding sides of similar triangles are in proportion.
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} \dots(1)$
Also,$\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \dots(2)$
Since $AD$ and $PM$ are medians,they divide their opposite sides into two equal parts.
$BD = \frac{BC}{2}$ and $QM = \frac{QR}{2} \dots(3)$
From equations $(1)$ and $(3),$ we obtain
$\frac{AB}{PQ} = \frac{BC/2}{QR/2} = \frac{BD}{QM} \dots(4)$
In $\Delta ABD$ and $\Delta PQM,$
$\angle B = \angle Q$ [Using equation $(2)$]
$\frac{AB}{PQ} = \frac{BD}{QM}$ [Using equation $(4)$]
$\therefore \Delta ABD \sim \Delta PQM$ (By $SAS$ similarity criterion)
$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$