Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\Delta ABC \sim \Delta PQR$.

(A) Given that,
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$
Let us extend $AD$ and $PM$ up to points $E$ and $L$ respectively,such that $AD = DE$ and $PM = ML$. Then,join $B$ to $E$,$C$ to $E$,$Q$ to $L$,and $R$ to $L$.
We know that medians divide opposite sides. Therefore,$BD = DC$ and $QM = MR$.
Also,$AD = DE$ (by construction) and $PM = ML$ (by construction).
In quadrilateral $ABEC$,diagonals $AE$ and $BC$ bisect each other at point $D$. Therefore,quadrilateral $ABEC$ is a parallelogram.
$\therefore AC = BE$ and $AB = EC$ (opposite sides of a parallelogram are equal).
Similarly,we can prove that quadrilateral $PQLR$ is a parallelogram and $PR = QL$,$PQ = LR$.
It was given that $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
$\Rightarrow \frac{AB}{PQ} = \frac{BE}{QL} = \frac{2AD}{2PM}$
$\Rightarrow \frac{AB}{PQ} = \frac{BE}{QL} = \frac{AE}{PL}$
$\therefore \Delta ABE \sim \Delta PQL$ (by $SSS$ similarity criterion).
We know that corresponding angles of similar triangles are equal.
$\therefore \angle BAE = \angle QPL \dots(1)$
Similarly,it can be proved that $\Delta AEC \sim \Delta PLR$ and $\angle CAE = \angle RPL \dots(2)$
Adding equation $(1)$ and $(2)$,we obtain:
$\angle BAE + \angle CAE = \angle QPL + \angle RPL$
$\Rightarrow \angle CAB = \angle RPQ \dots(3)$
In $\Delta ABC$ and $\Delta PQR$:
$\frac{AB}{PQ} = \frac{AC}{PR}$ (given)
$\angle CAB = \angle RPQ$ (using equation $(3)$)
$\therefore \Delta ABC \sim \Delta PQR$ (by $SAS$ similarity criterion).