Let ABCD is a square, AEB is an equilateral triangle described on the side of the square

and DBF is an equilateral triangle described on diagonal BD of  the square.

We have to prove that: Area(ΔDBF)/Area(ΔAEB) = 2/1

Proof:  If two equilateral triangles are similar then all angles are = 60 degrees.

Therefore, by AAA similarity criterion , △DBF ~ △AEB

Ar(ΔDBF)/Ar(ΔAEB) = DB² / AB²  ...............1

We know that the ratio of the areas of two similar triangles is equal to

the square of the ratio of their corresponding sides.

But, we have DB = √2AB .......2    {But diagonal of square is √2 times of its side}

Substitute equation (ii) in equation (i), we get

Ar(ΔDBF) / Ar(ΔAEB) = (√2AB )² / AB²   = 2AB² /AB² = 2/1

So, the area of equilateral triangle described on one side of square is equal to half the area of the equilateral triangle described

on one of its diagonals.