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

(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $\Delta ABC$ drawn to $BC$.
Since $AE$ is the altitude of an equilateral triangle,it bisects the base $BC$.
$\therefore BE = EC = \frac{BC}{2} = \frac{a}{2}$.
In the right-angled triangle $\Delta ABE$,by Pythagoras theorem:
$AE^{2} = AB^{2} - BE^{2} = a^{2} - (\frac{a}{2})^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3a^{2}}{4}$.
So,$AE = \frac{a\sqrt{3}}{2}$.
Given that $BD = \frac{1}{3} BC = \frac{a}{3}$.
Now,$DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{3a - 2a}{6} = \frac{a}{6}$.
In the right-angled triangle $\Delta ADE$,by Pythagoras theorem:
$AD^{2} = AE^{2} + DE^{2}$.
Substituting the values:
$AD^{2} = (\frac{a\sqrt{3}}{2})^{2} + (\frac{a}{6})^{2} = \frac{3a^{2}}{4} + \frac{a^{2}}{36}$.
Taking $LCM$ as $36$:
$AD^{2} = \frac{27a^{2} + a^{2}}{36} = \frac{28a^{2}}{36} = \frac{7a^{2}}{9}$.
Since $a = AB$,we have $AD^{2} = \frac{7 AB^{2}}{9}$.
Therefore,$9 AD^{2} = 7 AB^{2}$.
Hence proved.