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

If ΔABC and ΔDEF are so related that $\dfrac{AB}{FD} = \dfrac{BC}{DE} = \dfrac{CA}{EF}$, then which of the following is true?

1. ∠A = ∠E and ∠B = ∠D

2. ∠B = ∠F and ∠C = ∠D

3. ∠A = ∠F and ∠B = ∠D

4. ∠C = ∠F and ∠A = ∠D

In the given figure, PQ is parallel to TR, then by using condition of similarity:

1. $\dfrac{PQ}{RT} = \dfrac{OP}{OT} = \dfrac{OQ}{OR}$

2. $\dfrac{PQ}{RT} = \dfrac{OP}{OR} = \dfrac{OQ}{OT}$

3. $\dfrac{PQ}{RT} = \dfrac{OR}{OP} = \dfrac{OQ}{OT}$

4. $\dfrac{PQ}{RT} = \dfrac{OP}{OR} = \dfrac{OT}{OQ}$