In △PQR, MN is parallel to QR and $\dfrac{PM}{MQ} = \dfrac{2}{3}.$

(i) Find $\dfrac{MN}{QR}$.

(ii) Prove that △OMN and △ORQ are similar.

(iii) Find area of △OMN : area of △ORQ.

(i) Considering △PMN and △PQR,

∠ P = ∠ P (Common angles)
∠ PMN = ∠ PQR (Corresponding angles are equal)

Hence, by AA axiom △PMN ~ △PQR.

Given,

$\dfrac{PM}{MQ} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{PM}{PQ - PM} = \dfrac{2}{3} \\[1em] \Rightarrow 3PM = 2(PQ - PM) \\[1em] \Rightarrow 3PM = 2PQ - 2PM \\[1em] \Rightarrow 3PM + 2PM = 2PQ \\[1em] \Rightarrow 5PM = 2PQ \\[1em] \Rightarrow \dfrac{PM}{PQ} = \dfrac{2}{5}.$

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\therefore \dfrac{MN}{QR} = \dfrac{PM}{PQ} = \dfrac{2}{5}.$

Hence, $\dfrac{MN}{QR} = \dfrac{2}{5}$

(ii) Considering △OMN and △ORQ,

∠ MON = ∠ QOR (Vertically opposite angles are equal)
∠ OMN = ∠ ORQ (Alternate angles are equal)

Hence, by AA axiom △OMN ~ △ORQ.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \dfrac{MN^2}{QR^2} \\[1em] \Rightarrow \dfrac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \dfrac{2^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \dfrac{4}{25}.$

Hence, the ratio of the Area of △OMN : Area of △ORQ = 4 : 25.