Question -
Answer -
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD =1/3BC
Let the side of the equilateral trianglebe 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 Pythagorastheorem,
AD2 = AE2 + DE2
⇒ 9 AD2 = 7AB2