In an equilateral triangle ABC, D is a point on side BC such that BD = (1/3)BC. Prove that 9AD² = 7AB².

△ABC is an Equilateral triangle such that, AB=BC=AC
Construction: Draw an Altitude AE such that, E lies on BC and AE perpendicular to BC
In △AEB
By Pythagoras Theorem, AB² = AE² + EB² (1)
In △AED
By Pythagoras Theorem, AD² = AE² + ED² (2)
From 1) and 2)
AD² = ED² + AB² - EB² (3)
Since, EB = (1/2)BC = AB/2 and ED = BC/6 = AB/6
∴AD² = (AB/6)² + AB² - (AB/2)²
∴AD² = AB²/36 + AB² - 9AB²/36
∴AD² = 7AB²/36
∴9AD² = 7AB²