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

Given, Δ ABC ∼ Δ PQR

⇒ ∠ABC = ∠PQR (corresponding angles in similar triangle are equal)………..(1)

⇒ $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ (Ratio of corresponding sides of similar triangle are proportional)

⇒ $\dfrac{AB}{PQ} = \dfrac{\dfrac{BC}{2}}{\dfrac{QR}{2}}$

⇒ $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ (As D and M are mid-points of BC and QR) ………(2)

In Δ ABD and Δ PQM,

⇒ ∠ABD = ∠PQM [From (1)]

⇒ $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ [From (2)]

∴ Δ ABD ∼ Δ PQM (By S.A.S. axiom)

We know that,

Ratio of corresponding sides of similar triangle are similar.

$\therefore \dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$.

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