Step 1: Draw an equilateral triangle ABC and locate the given point (D) on side (BC).

Step 2: Draw AE ⊥ BC and join A and D.

          CONSTRUCTION:

Step 3: Locate the similar triangle in the figure.

            EXAMPLE: In a ΔABC and ΔACE

                                  AB = AC ( since, Equilateral triangle )

                                  AE = AE ( common side)

                           \angle AEB = \angle AEC = 90

                              ∴ ΔABC ≅ ΔACE ( For RHS criterion)

Step 4: Apply the Pythagoras theorem to the right triangles in the figure.

            EXAMPLE:   In a right angled triangle ADE

                                 AD^2 = AE^2 + DE^2 ..........................(1)

                               ∆ABE is also a right angle triangle

                                  ∴ AB^2 = BE^2 + AE^2 ...................(2)

Step 5: Subtract equation (1) from (2)

             From equation (1) and (2) ;

           AB^2 - AD^2 = BE^2 + AE^2 - AE^2 - DE^2

           AB^2 - AD^2 = BE^2 + AE^2 - AE^2 - DE^2

           AB^2 - AD^2 = (BE - DE)^2 ................(3)

Step 6: Write the above  equation in terms of BC and manipulate for the required equation.

             NOTE: BE = EC [By Corresponding parts of congruent triangles

                         are congruent" (CPCTC)]

                         BE = EC = \frac{BC}{2}

                         BD = \frac{1}{3} BC (GIVEN)

            EXAMPLE: From equation (3)

           AB^2 - AD^2 = (\frac{BC^2}{2} - DE)^2

                                  =  [math] [\frac{BC^2}{2} - (BE - BD)]^2 [/math] (Since DE = BE - BD)

                                  =  [math] [\frac{BC^2}{2} - (\frac{BC^2}{2} - \frac{1}{3} BC )]^2 [/math]  

                                  = \frac{BC^2}{4} - (\frac{BC^2}{2} - \frac{BC}{3})^2

                                  = \frac{BC^2}{4} - (\frac{3BC^2 - 2BC}{6})^2

                                  = \frac{BC^2}{4} - (\frac{BC^2}{36})

           AB^2 - AD^2 = \frac{8BC^2}{36}

           We know that AB = BC = CA

               So,   AB^2 = AD^2 + \frac{8AB^2}{36}

                       AB^2 = AD^2 + \frac{2AB^2}{9}

                       9AB^2 - 2AB^2 = 9AD^2

                       9AD^2 = 7AB^2