Sides AB, BC and median AD of the triangle ABC are respectively proportional to sides PQ, QR and median PM of triangle PQR.

Prove that:

(i) Δ ABD ∼ Δ PQM

(ii) Δ ABC ∼ Δ PQR

(i) Given, AB, BC and median AD of the triangle ABC are respectively proportional to sides PQ, QR and median PM of triangle PQR.

$\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{BC}}{\text{QR}} = \dfrac{\text{AD}}{\text{PM}}$

AD is median of triangle ABC.

∴ BD = DC = $\dfrac{1}{2}$ BC

PM is median of triangle PQR.

∴ QM = RM = $\dfrac{1}{2}$ QR

In Δ ABD and Δ PQM,

$\Rightarrow \dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{BC}}{\text{QR}} = \dfrac{\text{AD}}{\text{PM}}\\[1em] \Rightarrow \dfrac{\text{AB}}{\text{PQ}} = \dfrac{\dfrac{1}{2}\text{BC}}{\dfrac{1}{2}\text{QR}} = \dfrac{\text{AD}}{\text{PM}}\\[1em] \Rightarrow \dfrac{\text{AB}}{\text{PQ}} = \dfrac{BD}{QM} = \dfrac{\text{AD}}{\text{PM}}\\[1em]$

∴ Δ ABD ∼ Δ PQM (By SSS postulates)

Hence, Δ ABD ∼ Δ PQM.

(ii) From (i), Δ ABD ∼ Δ PQM

Corresponding angles of similar triangles are equal.

∴ ∠ABD = ∠PQM …………….(1)

In Δ ABC and Δ PQR,

⇒ ∠ABC = ∠PQR [From equation (1)]

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

∴ Δ ABC ∼ Δ PQR (By SAS postulate)

Hence, Δ ABC ∼ Δ PQR.