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06/06/2021 12:16 pm
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In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
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06/06/2021 12:22 pm
ABC is an equilateral triangle.
And D is a point on side BC such that BD = 1/3BC
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE, by Pythagoras theorem,
AD2 = AE2 + DE2
AD2 = \((\frac{a \sqrt{3}}{2})^2 + (\frac{a}{6})^2\)
= \((\frac{3a^2}{4} + (\frac{a^2}{36})\)
= \(\frac{28a^2}{36}\)
= \(\frac{7}{6}AB^2\)
⇒ 9 AD2 = 7 AB2