△ABC is an equilateral triangle in which D, E and F are the points on sides BC, AC and AB, respectively, such that AD ⊥ BC, BE ⊥ AC and CF ⊥ AB. Which of the following is true?

$3AC^2 = 4BE^2$

We know that,

Height of the equilateral triangle = \(\frac{\sqrt {3}}{2}\) × Side

Here,

Side = AB = BC = CA

Height = AD = BE = CF

\(\frac{\sqrt {3}}{2}\) × AC = BE

  \(\sqrt {3}\)  AC = 2BE

Squaring on both sides = 3 AC² = 4 BE²