Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of △ PQR. Show that △ ABC ~ △ PQR.

In Δ ABC and Δ PQR,

As, sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of △ PQR.

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$ …….(1)

AD and PM are median of Δ ABC and Δ PQR respectively.

$\therefore \dfrac{BD}{QM} = \dfrac{\dfrac{BC}{2}}{\dfrac{QR}{2}} = \dfrac{BC}{QR}.$

Substituting value of $\dfrac{BC}{QR}$ from above equation in (1), we get :

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

∴ Δ ABD ∼ Δ PQM [By SSS axiom]

In Δ ABC and Δ PQR,

⇒ $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ [Given]

⇒ ∠ABC = ∠PQR [∵ Δ ABD ∼ Δ PQM]

∴ Δ ABC ∼ Δ PQR [By S.A.S. axiom]

Hence, proved that Δ ABC ∼ Δ PQR.