Sides AB and AC and median AD of a △ 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 L such that $PM = ML$ and join LR.
$\therefore \triangle ABD \cong \triangle ECD$
$\therefore AB = EC$
Similarly, $PQ = LR$
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$
$\frac{EC}{LR} = \frac{AC}{PR} = \frac{2AD}{2PM} = \frac{AE}{PL}$
$$\begin{aligned}& \therefore \triangle AEC \sim \triangle PLR \\ & \Rightarrow \angle 2 = \angle 4\end{aligned}$$
Similarly, $\angle 1 = \angle 3$
Adding both, $\angle BAC = \angle QPR$
$\therefore \triangle ABC \sim \triangle PQR$