In an equilateral triangle ABC, D is a point on side BC such that BD = \frac{1}{3} BC. Prove that 9AD^2 = 7AB^2.

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