n an equilateral ABC, D is a point on side BC such that BD =??BC prove

1.	n an equilateral ABC, D is a point on side BC such that BD =??BC prove that 9AD2= 7AB
Answer» 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

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