Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ is a median in $\Delta ABC$ and $\overline{PM}$ is a median in $\Delta PQR.$ Prove that $\frac{AD}{PM} = \frac{AB}{PQ}.$

(N/A) Given: $\Delta ABC \sim \Delta PQR$ with correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ and $\overline{PM}$ are medians of $\Delta ABC$ and $\Delta PQR$ respectively.
To prove: $\frac{AD}{PM} = \frac{AB}{PQ}.$
Proof:
$1$. Since $\Delta ABC \sim \Delta PQR,$ the corresponding sides are proportional and corresponding angles are congruent.
$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle B \cong \angle Q \quad \dots (1)$
$2$. Since $\overline{AD}$ and $\overline{PM}$ are medians,$D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR.$
Therefore,$BC = 2BD$ and $QR = 2QM.$
$3$. Substituting these into equation $(1)$:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM} \quad \dots (2)$
$4$. In $\Delta ABD$ and $\Delta PQM$:
From $(1)$,$\angle B \cong \angle Q.$
From $(2)$,$\frac{AB}{PQ} = \frac{BD}{QM}.$
$5$. By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABD \sim \Delta PQM.$
$6$. Since the triangles are similar,their corresponding sides are proportional:
$\therefore \frac{AD}{PM} = \frac{AB}{PQ}.$ Hence proved.