Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

(N/A) Let us assume two similar triangles $\Delta ABC \sim \Delta PQR$. Let $AD$ and $PS$ be the medians of these triangles.
Since $\Delta ABC \sim \Delta PQR$,we have:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \quad ...(1)$
And $\angle B = \angle Q \quad ...(2)$
Since $AD$ and $PS$ are medians,$D$ and $S$ are midpoints of $BC$ and $QR$ respectively.
Therefore,$BD = \frac{BC}{2}$ and $QS = \frac{QR}{2}$.
Substituting these into equation $(1)$:
$\frac{AB}{PQ} = \frac{2BD}{2QS} = \frac{BD}{QS} \quad ...(3)$
In $\triangle ABD$ and $\triangle PQS$:
$\angle B = \angle Q$ [From $(2)$]
$\frac{AB}{PQ} = \frac{BD}{QS}$ [From $(3)$]
Therefore,$\triangle ABD \sim \triangle PQS$ by $SAS$ similarity criterion.
This implies that $\frac{AB}{PQ} = \frac{AD}{PS} \quad ...(4)$
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2$
From equation $(4)$,substituting $\frac{AB}{PQ} = \frac{AD}{PS}$:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AD}{PS}\right)^2$
Hence proved.