If ∆ABC and ∆BDE are equilateral triangles,

Given: In ΔABC and ΔBDE are equilateral triangles. D is the midpoint of BC.

To find: Ar∆ABCAr∆BDE

In ΔABC and ΔBDE

$\triangle \mathrm{ABC} \sim \triangle \mathrm{BDE}(\mathrm{AAAcrietria}$ of similarity, all angles of equilateral triangle are equal $)$

Since D is the midpoint of BC, BD : DC = 1.

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

Let DC = x, and BD = x

Therefore BC = BD + DC = 2x

Hence 

$\frac{A r(\Delta \mathrm{ABC})}{A r(\Delta \mathrm{BDE})}=\frac{\mathrm{BC}^{2}}{\mathrm{BD}^{2}}$

$=\frac{(\mathrm{BD}+\mathrm{DC})^{2}}{(\mathrm{BD})^{2}}$

$=\frac{(1 x+1 x)^{2}}{(1 x)^{2}}$

$=\frac{(2 x)^{2}}{(1 x)^{2}}$

$\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\triangle \mathrm{BDE})}=\frac{4}{1}$