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

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

Question 333565: Given: Equilateral triangle ABC
Let D be a point on BC such that BD = BC
Prove that 9*AD� = 7*AB�
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!


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

Prove that 9*AD� = 7*AB� 



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:

AD� = AB� + BD� - 2(AB)(BD)cos(B)

 x� = 3� + 1� - 2(3)(1)cos(60�)

 x� = 9 + 1 - 6(.5)

 x� = 10 - 3

 x� = 7 

AD� = 7

9*AD� = 9*7 = 63

AB = 3

7*AB� = 7*3^2 = 7*9 = 63.

So 9*AD� = 7*AB� since both = 63.

Edwin

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