Question

16. In an equilateral triangle ABC, D is a point on side BC such that BD = 3 BC. Prove that
9 ADP=7 AB?
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Answer 1

Answer:

it's the proved part

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Answer 2

Answer:

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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