<s>[INST] Equilateral triangle trisected In an equilateral triangle $ABC$ the side $BC$ is trisected by points $D$ and $E$. Prove $9|AD|^2 = 7|AB|^2$.
 [/INST] 
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = $\frac{BC}{2}$ = $\frac{a}{2}$
And using phythagorous theorem, 
AE = $\sqrt{AB^2 - BE^2}$
AE = $\frac{\sqrt{3}a}{2}$
Given that, BD = $\frac{1}{3}$ BC
∴ BD = $\frac{a}{3}$
DE = BE - BD =  $\frac a{2} - \frac a{3} = \frac a{6}$
Applying Pythagoras theorem in ΔADE, we obtain
$AD^2 = AE^2 + DE^2$
= $\left(\frac{\sqrt{3}a}{2}\right)^2 + \left(\frac{a}{6}\right)^2$
=$\left(\frac{3a^2}{4}\right) + \left(\frac{a^2}{36}\right)$
$AD^2 = \frac{28}{36} a^2$
$9 AD^2 = 7 AB^2$
</s>