In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3} \) BC. Prove that \(9AD^2 = 7AB^2\).

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = \(\frac{1}{3} \) ×BC



Let the side of the equilateral triangle be a, and AE be the altitude of \(\triangle\) ABC.

BE = EC = \(\frac{BC}{2} \) = \(\frac{a}{2} \)
And, \(AE = {{a \sqrt{3}} \over {2}}\)
Given, BD = \(\frac{1}{3} \) ×BC
\(\Rightarrow \) BD = \(\frac{a}{3} \)

\(\therefore \) DE = BE – BD
= \(\frac{a}{2} – \frac{a}{3} = \frac{a}{6} \)

In \(\triangle\) ADE, by Pythagoras theorem, \(AD^2 = AE^2 + DE^2\) \(AD^2 = ({{a\ sqrt{3}} \over {2}})^2 + ({{a} \over {6}})^2\)
\(= ({{3a^2} \over {4}}) + ({{a^2} \over {36}})\)
\(={{28a^2} \over {36}}\)
\(={{7} \over {9}} AB^2\)
\(9 AD^2 = 7 AB^2\)