In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
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 = 1/3BC
∴ BD = a/3
DE = BE - BD = a/2 - a/3 = a/6
Applying Pythagoras theorem in ΔADE, we get
AD2 = AE2 + DE2
⇒ 9 AD2 = 7 AB2
