In an eqilateral triangle ABCD is …

Given,  {tex}\triangle ABC{/tex} in which {tex}AB = BC = CA{/tex} and D is a point on BC such that BD ={tex}\frac { 1 } { 3 }{/tex}BC.
Prove: 9AD² = 7AB².
Construction :Draw {tex}A L \perp B C{/tex} 
Proof: In right triangles ALB and ALC, we have
{tex}AB = AC {/tex}(given)
{tex}AL = AL {/tex}(common)
{tex}\therefore \triangle A L B \cong \triangle A L C{/tex} [by RHS axiom]
So, {tex}BL = CL.{/tex}
Thus, BD ={tex}\frac { 1 } { 3 }{/tex}BC and BL ={tex}\frac { 1 } { 2 }{/tex}BC.
In {tex}\triangle ALB{/tex}, {tex}\angle A L B = 90 ^ { \circ }{/tex}
By using pythagoras theorem, we get 
{tex}\therefore{/tex} AB² = AL² + BL² ...(i)
In {tex}\triangle ALD{/tex}, {tex}\angle A L D = 90 ^ { \circ }{/tex}
By using pythagoras theorem, we get 
{tex}\therefore{/tex} AD² = AL² + DL² 
= AL² + (BL - BD)²
= AL² + BL² + BD² - {tex}2 B L \cdot B D{/tex}
= (AL² + BL²) + BD² - {tex}2 B L \cdot B D{/tex}
= AB² + BD² - {tex}2 B L \cdot B D{/tex} [using (i)]
{tex}= B C ^ { 2 } + \left( \frac { 1 } { 3 } B C \right) ^ { 2 } - 2 \left( \frac { 1 } { 2 } B C \right) \cdot \frac { 1 } { 3 } B C{/tex}
{tex}= B C ^ { 2 } + \frac { 1 } { 9 } B C ^ { 2 } - \frac { 1 } { 3 } B C ^ { 2 }{/tex}
{tex}= \frac { 7 } { 9 } B C ^ { 2 } = \frac { 7 } { 9 } A B ^ { 2 }{/tex} [{tex}\because{/tex} BC = AB]
Therefore, 9AD² = 7AB².