Precal Question

See the following image :

Since AD bisects BAC   then 

BA /CA  =  BD/DC

3 / 6   =  BD /DC

1 /2   =  BD/DC

DC = 2BD

BC = BD + DC

BC = BD + 2BD

BC = 3BD

So  BD =  (1/3)BC =  (1/3)(8)  = 8/3

And  DC = 2BD = 2(8/3)  =  16/3

So....using the Law of Cosines we have

(8/3)^2  =  AD^2 + 3*2 - 2(AD)(3) cosBAD   (1)

(16/3)^2  = AD^2 +6^2  - 2(AD)(6) cosDAC   (2)

And since AD  bisects BAC....then cos  BAD  =  cos DAC

So.....

(64/9) - 9  = AD^2 - 6(AD)cosBAD     (3)

(256/9) - 36  = AD^2 - 12AD cosBAD  (4)

-17/9  = AD^2 - 6(AD) cosBAD   (5)

-68/9  = AD^2 - 12(AD) cosBAD    (6)   

Multiply (5)  by -2

34/9  = -2AD^2 + 12(AD)cosBAD   (7)

-68/9  =  AD^2  - 12(AD)cosBAD    (8)       add (7) and (8)

-34/9 = -AD^2   

34/9  = AD^2

√34/3 = AD

Therefore....using (5)  and subbing for AD....  we have that

-17/9 = 34/9 - 6(√34/3)cosBAD

-51/9  = - 2√34cosBAD

cos BAD  =  (51) / ( 9 * 2√34)

cos BAD  = (51) (18√34)

cos BAD  =  17/ (6√34)  =  (17√34) /  (6 * 34)  =  √34 / 12

In triangle ABC, AB = 3, AC = 6, BC = 8, and D lies on BC such that AD bisects ∠BAC. Find cos ∠BAD.

$\begin{array}{|rcll|} \hline \mathbf{\cos(\angle BAD)} &=& \mathbf{\sqrt{\dfrac{s\cdot (s-BC)}{AB\cdot AC} }} \quad | \quad \mathbf{s = \dfrac{AB+AC+BC}{2}}=\dfrac{3+6+8}{2}=8.5 \\\\ &=& \sqrt{\dfrac{8.5\cdot (8.5-8)}{3\cdot 6} } \\\\ &=& \sqrt{\dfrac{8.5\cdot (0.5)}{18} } \\\\ &=& \sqrt{\dfrac{4.25}{18} } \\\\ &=& \sqrt{\dfrac{2.125}{9} } \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ \mathbf{\cos(\angle BAD)} &=& \mathbf{ 0.48591265790 } \\ \hline \end{array}$