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Question
In an equilateral triangle ABC, D is a point on side BC such that BD = `1/3BC` . Prove that 9 AD2 = 7 AB2
Solution
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
`∴ BE = EC = (BC)/2 = a/2`
And, AE = `(asqrt3)/2`
Given that, BD = `1/3BC`
∴ BD = a/3
`DE = BE - BD = a/2 - a/3 = a/6`
Applying Pythagoras theorem in ΔADE, we get
AD2 = AE2 + DE2
`AD^2 = ((asqrt3)/2)^2 + (a/6)^2`
`= ((3a^2)/4) + (a^2/36)`
`= (28a^2)/36`
`= 7/9 AB^2`
⇒ 9 AD2 = 7 AB2
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