$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. $\overline{AD}$ and $\overline{PM}$ are medians of these triangles. Prove that $AB \times PM = PQ \times AD$.

(A) Given: $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,their corresponding sides are proportional and their corresponding angles are equal.
Therefore,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle B = \angle Q$.
Since $\overline{AD}$ and $\overline{PM}$ are medians,$D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR$.
Thus,$BC = 2BD$ and $QR = 2QM$.
Substituting these into the ratio: $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
Now,in $\Delta ABD$ and $\Delta PQM$:
$1$. $\frac{AB}{PQ} = \frac{BD}{QM}$ (Proved above)
$2$. $\angle B = \angle Q$ (Corresponding angles of similar triangles)
By $SAS$ similarity criterion,$\Delta ABD \sim \Delta PQM$.
Since the triangles are similar,the ratio of their corresponding sides is equal: $\frac{AB}{PQ} = \frac{AD}{PM}$.
By cross-multiplication,we get $AB \times PM = PQ \times AD$.
Hence proved.