SOLUTION: Given: Equilateral triangle ABC Let D be a point on BC such that BD = {{{1/3}}}BC Prove that 9*AD² = 7*AB²

Given: Equilateral triangle ABC 
Let D be a point on BC such that BD = BC

Prove that 9*AD² = 7*AB² 



Let BD be 1 unit long, then BC is 3 units long which
means that DC is 2 units long, BD = BC.
And since triangle ABC is equilateral, 
AB = AC = BC = 3 and angle B = 60°
Let AD = x.



I am going to assume that you have had the law of cosines.  If
you have not studied that yet, then post again, or email me
and I will show you how to do it using only the Pythagorean
theorem.

Using the law of cosines on triangle ABD:

AD² = AB² + BD² - 2(AB)(BD)cos(B)

 x² = 3² + 1² - 2(3)(1)cos(60°)

 x² = 9 + 1 - 6(.5)

 x² = 10 - 3

 x² = 7 

AD² = 7

9*AD² = 9*7 = 63

AB = 3

7*AB² = 7*3^2 = 7*9 = 63.

So 9*AD² = 7*AB² since both = 63.

Edwin