\(\Delta ABC\) is an equilateral triangle. Point D is on side BC such that \(BD = \dfrac{1}{3} BC\). Prove that \(9AD^2 = 7AB^2\).

Answer: \(9AD^2 = 7AB^2\) (proved below)

[Figure: equilateral \(\Delta ABC\); D on BC with BD one-third of BC; AM \(\perp\) BC.]

Let each side be \(a\). Draw AM \(\perp\) BC. In an equilateral triangle, M is the midpoint of BC: \(BM = \dfrac{a}{2}\), and \(AM = \dfrac{\sqrt{3}}{2}a\).

\(BD = \dfrac{a}{3}\), so \(DM = BM - BD = \dfrac{a}{2} - \dfrac{a}{3} = \dfrac{a}{6}\).

In right-angled \(\Delta AMD\): \(AD^2 = AM^2 + DM^2 = \dfrac{3a^2}{4} + \dfrac{a^2}{36} = \dfrac{27a^2 + a^2}{36} = \dfrac{28a^2}{36} = \dfrac{7a^2}{9}\).

\(\therefore 9AD^2 = 7a^2 = 7AB^2\). Hence proved.