In equuialteral triangle ABC, D is the point on side BC such that BD =1/3BC prove that 9(AD)²=7(AB)²
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Construction: Draw AP ⟂ BC
In Δ ADP,
AD2 = AP2 + DP2
AD2 = AP2 + (BP - BD)2
AD2 = AP2 + BP2 + BD2 - 2(BP)(BD)
AD2 = AB2 + (1/3 BC)2 - 2(BC/2)(BC/3)
AD2 = 7/9 AB2 (Since BC = AB)
9AD2 = 7AB2
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