In an equilateral ABC, D is a point on side BC such that  Prove that 9AD² = 7AB². 

Const: Draw AE ⊥ BC.
Proof: In right triangles ABE and ACE, we have
AE = AE    [common]
∠AEB = ∠AEC    [90°]
and    AB = AC
[∆ABC is an equilateral]
Therefore, by using RHS congruent condition
                                       
                                          BE = CE         [by CPCT]
                                          
In right triangle ADE, we have
                          
                     
                  
                     
                            


  
             
But             AB = BC = CA