ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .Now, ∆ABE and ∆AEC∠AEB = ∠ACE = 90°AE is common side of both triangles ,AB = AC [ all sides of equilateral triangle are equal ]From R – H – S congruence rule ,∆ABE ≡ ∆ACE∴ BE = EC = BC/2Now, from Pythagoras theorem,∆ADE is right angle triangle ∴ AD² = AE² + DE² ——(1)∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ——(2)From equation (1) and (2)AB² – AD² = BE² – DE²= (BC/2)² – (BE – BD)²= BC²/4 – {(BC/2) – (BC/3)}²= BC²/4 – (BC/6)²= BC²/4 – BC²/36 = 8BC²/36 = 2BC²/9∵AB = BC = CASo, AB² = AD² + 2AB²/99AB² – 2AB² = 9AD²Hence, 9AD² = 7AB²