Triangles ABC and PQR are similar to each other, then :

1. $\dfrac{\text{Ar.(△ABC)}}{\text{Ar.(△PQR)}} = \dfrac{BC^2}{PQ^2}$

2. $\dfrac{AB^2}{PQ^2} = \dfrac{AC^2}{PR^2}$

3. $\dfrac{\text{Ar.(△BAC)}}{\text{Ar.(△QPR)}} \ne \dfrac{AB^2}{QP^2}$

4. $\dfrac{AC^2}{PR^2} = \dfrac{BC^2}{PQ^2}$

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$\Rightarrow \dfrac{\text{Ar.(△ABC)}}{\text{Ar.(△PQR)}} = \dfrac{AB^2}{PQ^2}$ …….(1)

$\Rightarrow \dfrac{\text{Ar.(△ABC)}}{\text{Ar.(△PQR)}} = \dfrac{AC^2}{PR^2}$ ………(2)

From equation (1) and (2), we get :

$\Rightarrow \dfrac{AB^2}{PQ^2} = \dfrac{AC^2}{PR^2}$

Hence, Option 2 is the correct option.