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

Given,

The two triangles are ΔABC and ΔPQR.

$\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$

By SSS similarity criterion, if the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal and the triangles are similar.

To find the correspondence of vertices:

The side AB corresponds to QR. The vertex opposite to AB is C, and opposite to QR is P.

C = P

The side BC corresponds to PR. The vertex opposite to BC is A, and opposite to PR is Q.

A = Q

The side CA corresponds to PQ. The vertex opposite to CA is B, and opposite to PQ is R.

B = R

∴ΔPQR ∼ ΔCAB

Hence, option 1 is the correct option.