Δ ABC is an equilateral triangle

By Pythagoras theorem in triangle ABD

AB² = AD² + BD²

but BD = 1/2 BC (∵ In a triangle, the perpendicular from the vertex to the base bisects the base)

thus AB² = AD² + {1/2 BC}²

AB² = AD² + 1/4 BC²

4 AB² = 4AD² + BC²

4 AB² - BC² = 4 AD²

(as AB = BC we can subtract them )

Thus 3AB² = 4AD²