 Krishna
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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

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

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

= \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})