If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is

1. 2 : 1

2. 1 : 2

3. 1 : 4

4. 4 : 1

Since triangles ABC and BDE are equilateral triangles so, each angle will be equal to 60°.

Since all angles are equal to 60°.

Hence, by AAA axiom △ABC ~ △BDE.

Since D is the midpoint of BC so,

$\Rightarrow BC = 2BD \\[1em] \Rightarrow \dfrac{BC}{BD} = \dfrac{2}{1}.$

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △BDE}} = \dfrac{BC^2}{BD^2} \\[1em] = \dfrac{2^2}{1^2} \\[1em] = \dfrac{4}{1} \\[1em] = 4 : 1.$

Hence, Option 4 is the correct option.