Question 15:

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD² = 7 AB².

Answer:

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 that, BD = 1/3 BC

∴ BD = a/3

DE = BE − BD = (a/2) – (a/3) = a/6

Applying Pythagoras theorem in ΔADE, we obtain

AD² = AE² + DE²

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