In an equilateral triangle ABC,D is …

We have {tex}BD = \frac{1}{3}BC{/tex} Draw AP ​{tex} \bot {/tex}​ BC
In ​{tex}\triangle {/tex}​APB,
AB² = AP² + BP² = AP² + (BD + DP)²
= AP² + BD² + DP² + 2BD.DP
= (AP² + DP²) + BD² + 2BD.DP
=AD² + DB² + 2BD.DP [AP² + DP² = AD²]
 
=AD²+{tex}{\left( {\frac{1}{3}BC} \right)^2} + 2\left( {\frac{1}{3}BC} \right)(BP - BD){/tex}
={tex}AD^2 + \frac{1}{9}BC + \frac{2}{3}BC\left( {\frac{1}{2}BC - \frac{1}{3}BC} \right){/tex} {tex}\left[ {BP = \frac{1}{2}BC,BD = \frac{1}{3}BC} \right]{/tex}
={tex}A{D^2} + \frac{1}{9}A{B^2} + \frac{2}{3}AB\left( {\frac{1}{2}AB - \frac{1}{3}AB} \right){/tex} [BC = AB, Sides of an equilateral{tex}\vartriangle {/tex}​ ]
={tex}A{D^2} + \frac{1}{9}A{B^2} + \frac{1}{3}A{B^2} - \frac{2}{9}A{B^2}{/tex}
={tex}A{D^2} + \frac{1}{9}A{B^2} + \frac{1}{9}A{B^2}{/tex}
​{tex}\Rightarrow {/tex}​ {tex}A{B^2} = A{D^2} + \frac{2}{9}A{B^2}{/tex}
​{tex}\Rightarrow {/tex}​ 9AB² = 9AD² + 2AB²
​{tex}\Rightarrow {/tex} ​9AB² = 9AB² + 2AD²
​{tex}\Rightarrow {/tex}​ 7AB² = 9AD²
​{tex}\Rightarrow {/tex}​ 9AD² = 7AB² Proved