If AD and PM are medians of triangles ABC and PQR, respectively where ∆ ABC ~ ∆ PQR, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.

Given, ΔABC ∼ ΔPQR

⇒ ∠ABC = ∠PQR (corresponding angles) --------- (1)

⇒ AB/PQ = BC/QR (corresponding sides)

⇒ AB/PQ = (BC/2) / (QR/2)

⇒ AB/PQ = BD/QM (D and M are mid-points of BC and QR) ------------ (2)

In ΔABD and ΔPQM,

∠ABD = ∠PQM (from 1)

AB/PQ = BD/QM (from 2)

⇒ ΔABD ∼ ΔPQM (SAS criterion)

⇒ AB/PQ = BD/QM = AD/PM (corresponding sides)

⇒ AB/PQ = AD/PM

Hence proved.