In ΔPQR, MN is parallel to QR and $\dfrac{PM}{QM} = \dfrac{2}{3}$.

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

(ii) Prove that ΔOMN and ΔORQ are similar.

(iii) Find: Area of ΔOMN : Area of ΔORQ.

(i) Considering ΔPMN and ΔPQR,

∠P = ∠P [Common angles]

∠PMN = ∠PQR [Corresponding angles are equal]

∴ ΔPMN ∼ ΔPQR by AA similarity.

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,

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

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

(ii) Considering ΔOMN and ΔORQ,

∠MON = ∠QOR (Vertically opposite angles are equal)

∠OMN = ∠ORQ (Alternate angles are equal)

Hence, by AA similarity ΔOMN ∼ ΔORQ.

(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.

$\therefore \dfrac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \dfrac{MN^2}{QR^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \dfrac{2^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \dfrac{4}{25} \\[1em]$

Hence, the ratio of the Area of ΔOMN : Area of ΔORQ = 4 : 25.