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Answer :
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,
\(AD^2 = AE^2 + DE^2\)
\(AD^2 = ({{a\ sqrt{3}} \over {2}})^2 + ({{a} \over {6}})^2\)
\(= ({{3a^2} \over {4}}) + ({{a^2} \over {36}})\)
\(={{28a^2} \over {36}}\)
\(={{7} \over {9}} AB^2\)
\(9 AD^2 = 7 AB^2\)