Question

In an equilateral triangle ABC, D is a point on side BC such that BD =1/3 BC . Prove that 9AD^2=7AB^2.

Answer 1

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Answer 2

in triangle ADP

AD^2 = AP^2 + DP^2

AD^2 = AP^2 + (BP-BD)^2

AD^2 = AP^2 + BP^2 + BD ^2 - 2BP.BD

AD^2 = AP^2 + (1/3 BC^2 ) -2 (BC/2) . (BC/3)

AD^2 = 7/9 AB^2 { SINCE BC=AB}

9AD^2 = 7 AB^2

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