Asked in: July 2024 board paper\(\Delta ABC\) is an equilateral triangle. Point D is on seg BC such that \(BD = \dfrac{1}{5} BC\). Then prove that \(\dfrac{AD^2}{AB^2} = \dfrac{21}{25}\).

Answer: \(\dfrac{AD^2}{AB^2} = \dfrac{21}{25}\) (proved below)

[Figure: equilateral \(\Delta ABC\); D on BC near B; AM is the perpendicular from A to BC.]

Let each side be \(a\): AB = BC = AC = \(a\). Then \(BD = \dfrac{a}{5}\).

Construction: draw seg AM \(\perp\) BC. In an equilateral triangle the perpendicular bisects the base, so \(BM = \dfrac{a}{2}\), and \(AM = \dfrac{\sqrt{3}}{2}a\) (30°–60°–90° triangle ABM).

D lies between B and M because \(\dfrac{a}{5} \lt \dfrac{a}{2}\). So \(DM = BM - BD = \dfrac{a}{2} - \dfrac{a}{5} = \dfrac{3a}{10}\).

In right-angled \(\Delta AMD\), by Pythagoras theorem: \(AD^2 = AM^2 + DM^2 = \dfrac{3a^2}{4} + \dfrac{9a^2}{100} = \dfrac{75a^2 + 9a^2}{100} = \dfrac{84a^2}{100} = \dfrac{21a^2}{25}\).

\(\therefore \dfrac{AD^2}{AB^2} = \dfrac{21a^2/25}{a^2} = \dfrac{21}{25}\). Hence proved.