Sides AB and AC and median AD to △ ABC are respectively proportional to sides PQ and PR and median PM of another…

Correct figure
Produce $AD$ to $E$ such that $AD = DE$ and join $EC$
Produce $PM$ to $N$ such that $PM = MN$ and join $NR$
$\triangle ADB \cong \triangle EDC$
$\therefore AB = EC$
Similarly, $PQ=NR$
Since, $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$
$\Rightarrow \frac{EC}{NR} = \frac{AC}{PR} = \frac{AE}{PN}$
$\therefore \triangle AEC \sim \triangle PNR$
$\Rightarrow \angle 1 = \angle 2$
Similarly, $\angle 3 = \angle 4$
Hence $\angle 1 + \angle 3 = \angle 2 + \angle 4$ or $\angle A = \angle P$
Also, $\frac{AB}{PQ} = \frac{AC}{PR}$
$\therefore \triangle ABC \sim \triangle PQR$