Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of △…

In $\triangle ABC$ and $\triangle PQR$
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$
$\frac{AB}{PQ} = \frac{2 BD}{2 QM} = \frac{AD}{PM}$
($\therefore$ D is midpoint of BC and M is midpoint of QR)
$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \Rightarrow \triangle ABD \sim \triangle PQM$
$\Rightarrow \angle B = \angle Q$ -(i)
Now, In $\triangle ABC$ and $\triangle PQR$
$\frac{AB}{PQ} = \frac{BC}{QR}$ (given)
$\angle B = \angle Q$ from (i)
$\therefore \triangle ABC \sim \triangle PQR$