# Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilatral triangle described on one of its diagonals

Arun
25757 Points
5 years ago
Dear Nicky
Here 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.
To Prove:    Ar(ΔDBF) / Ar(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   --------------------(i)
We know that the ratio of the areas of two similar triangles is equal to
the square of the ratio of their corresponding sides i .e.
But, we have DB = √2AB     {But diagonal of square is √2 times of its side} -----(ii).
Substitute equation (ii) in equation (i), we get
Ar(ΔDBF) / Ar(ΔAEB) = (√2AB )2 / AB2   = 2 AB2 / AB2 = 2
∴ Area of equilateral triangle described on one side os square is equal to half the area of the equilateral triangle described on one of its diagonals.
Regards