Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\Delta PQR$ (see Figure). Show that $\Delta ABC \sim \Delta PQR$.

(N/A) Given: In $\Delta ABC$ and $\Delta PQR$,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
Since $AD$ and $PM$ are medians,$D$ and $M$ are midpoints of $BC$ and $QR$ respectively.
Therefore,$BD = \frac{BC}{2}$ and $QM = \frac{QR}{2}$.
Substituting these in the given ratio:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
In $\Delta ABD$ and $\Delta PQM$:
$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$ (Proved above).
Therefore,$\Delta ABD \sim \Delta PQM$ by $SSS$ similarity criterion.
This implies $\angle B = \angle Q$ (Corresponding angles of similar triangles).
Now,in $\Delta ABC$ and $\Delta PQR$:
$1$. $\frac{AB}{PQ} = \frac{BC}{QR}$ (Given)
$2$. $\angle B = \angle Q$ (Proved above)
Therefore,$\Delta ABC \sim \Delta PQR$ by $SAS$ similarity criterion.