Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: $\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}$.

Given, ∆ABC ~ ∆PQR

AD and PM are the medians, so BD = DC and QM = MR

Since, corresponding sides of similar triangles are proportional.

$\dfrac{AB}{PQ} = \dfrac{BC}{QR}$

We can write,

$\dfrac{AB}{PQ} = \dfrac{\dfrac{BC}{2}}{\dfrac{QR}{2}} = \dfrac{BD}{QM}$

And, ∠ABC = ∠PQR i.e., ∠ABD = ∠PQM

∴ ∆ABD ~ ∆PQM [By SAS]

Since, corresponding sides of similar triangles are proportional.

$\dfrac{AB}{PQ} = \dfrac{AD}{PM}$.

Hence, proved that $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$.