In an equilateral triangle ABC, D …

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/2 Now, 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 = CA So, AB² = AD² + 2AB²/9 9AB² - 2AB² = 9AD² Hence, 9AD² = 7AB²?

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