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.

PQR is a triangle, S is a point on the side QR of ΔPQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

(i) Prove ΔPQR ∼ ΔSPR.

(ii) Find the length of QR and PS.

(iii) Find $\dfrac{\text{area of ΔPQR}}{\text{area of ΔSPR}}$.

If ΔABC ∼ ΔQRP, then the corresponding proportional sides are :

1. $\dfrac{AB}{PQ} = \dfrac{BC}{RP}$

2. $\dfrac{AB}{QR} = \dfrac{BC}{QP}$

3. $\dfrac{AC}{QR} = \dfrac{BC}{RP}$

4. $\dfrac{AB}{QR} = \dfrac{BC}{RP}$

If in ΔABC and ΔPQR, we have $\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$, then:

1. ΔPQR ∼ ΔCAB

2. ΔPQR ∼ ΔABC

3. ΔBCA ∼ ΔPQR

4. ΔCBA ∼ ΔPQR