\(\Delta ABC\) is an equilateral triangle. Point D is on seg BC such that BD \(= \dfrac{1}{5}\) BC. Then prove that \(\dfrac{AD^2}{AB^2} = \dfrac{21}{25}\).[Figure: equilateral triangle ABC with D on side BC, closer to B; seg AD is drawn, and AE is d…

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

Let AB = BC = AC = \(x\). Then BD \(= \dfrac{x}{5}\) and DC \(= \dfrac{4x}{5}\).

Draw AE \(\perp\) BC. In an equilateral triangle the altitude bisects the base, so E is the midpoint of BC: BE \(= \dfrac{x}{2}\), and \(AE^2 = AB^2 - BE^2 = x^2 - \dfrac{x^2}{4} = \dfrac{3x^2}{4}\).

\(DE = BE - BD = \dfrac{x}{2} - \dfrac{x}{5} = \dfrac{3x}{10}\).

In right \(\Delta AED\): \(AD^2 = AE^2 + DE^2 = \dfrac{3x^2}{4} + \dfrac{9x^2}{100} = \dfrac{75x^2 + 9x^2}{100} = \dfrac{84x^2}{100} = \dfrac{21x^2}{25}\).

Dividing by \(AB^2 = x^2\): \(\dfrac{AD^2}{AB^2} = \dfrac{21}{25}\).