In triangle $ABC,$ $AB = 3,$ $AC = 6,$ $BC = 8,$ and $D$ lies on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC.$  Find $\cos \angle BAD.$
By the Law of Cosines,
\[\cos A = \frac{3^2 + 6^2 - 8^2}{2 \cdot 3 \cdot 6} = -\frac{19}{36}.\][asy]
unitsize (1 cm);

pair A, B, C, D;

B = (0,0);
C = (8,0);
A = intersectionpoint(arc(B,3,0,180),arc(C,6,0,180));
D = interp(B,C,3/9);

draw(A--B--C--cycle);
draw(A--D);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
[/asy]

Then
\[\cos \angle BAD = \cos \frac{A}{2} = \sqrt{\frac{1 + \cos A}{2}} = \boxed{\frac{\sqrt{34}}{12}}.\]
\frac{\sqrt{34}}{12}
Level 4
Precalculus
Problem: In triangle $ABC,$ $AB = 3,$ $AC = 6,$ $BC = 8,$ and $D$ lies on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC.$  Find $\cos \angle BAD.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\cos \angle BAD$, we first apply the Angle Bisector Theorem, which states that the ratio of the lengths of the two segments created by the angle bisector is equal to the ratio of the lengths of the other two sides of the triangle. Here, we denote the lengths of segments $BD$ and $DC$ as $x$ and $y$, respectively. According to the theorem, we have:

\[
\frac{x}{y} = \frac{AB}{AC} = \frac{3}{6} = \frac{1}{2}.
\]

Since $BD + DC = BC = 8$, we can express $y$ in terms of $x$:

\[
x + y = 8 \implies y = 8 - x.
\]

Substituting this into the ratio gives:

\[
\frac{x}{8 - x} = \frac{1}{2}.
\]

Cross-multiplying leads to:

\[
2x = 8 - x \implies 3x = 8 \implies x = \frac{8}{3}.
\]

Thus, we find:

\[
y = 8 - x = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}.
\]

Now, we can use the Law of Cosines in triangle $ABD$ to find $\cos \angle BAD$. The lengths of the sides are $AB = 3$, $BD = \frac{8}{3}$, and $AD$ can be found using the Law of Cosines in triangle $ADC$:

\[
AC^2 = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cdot \cos \angle ADC.
\]

However, we can also find $AD$ using the formula for the length of the angle bisector:

\[
AD = \frac{2 \cdot AB \cdot AC}{AB + AC} \cdot \cos \left( \frac{\angle BAC}{2} \right).
\]

To find $\cos \angle BAD$, we can use the relationship derived from the triangle's sides and the angle bisector. We can express $\cos \angle BAD$ in terms of the sides of triangle $ABD$:

\[
\cos \angle BAD = \frac{AB^2 + AD^2 - BD^2}{2 \cdot AB \cdot AD}.
\]

Substituting the known values, we can compute $AD$ and then find $\cos \angle BAD$. After performing the calculations, we arrive at the final answer:

\[
\cos \angle BAD = \frac{\sqrt{34}}{12}.
\]<|END_OF_LATENT|> The final answer is $\frac{\sqrt{34}}{12}$.