Given that ABC is an equilateral triangle. Therefore, its sides and angles are equal.
Hence, AB = BC = AC and \(\angle\)ABC = \(\angle\)ACB = \(\angle\)BAC = 60°.
Draw an altitude CD from C to the side AB.
Now, in right angle triangle ∆BDC,
⇒ 3AB2 = 4CD2 which implies that three times the square of one side is equal to four times the square of one of its altitude.