### Instruction: We have a triangle $\triangle ABC$ such that $AB = 6,$ $BC = 8,$ and $CA = 10.$ If $AD$ is an angle bisector such that $D$ is on $BC,$ then find the value of $AD^2.$.### Response: I know that angle bisectors in a triangle divide the opposite side in the same ratio as the adjacent sides.
So, if $AD$ is an angle bisector, then $BD/DC = AB/AC.$
I can plug in the given values to get $BD/DC = 6/10 = 3/5.$
This means that $BD = 3k$ and $DC = 5k$ for some constant $k.$
Since $BC = 8,$ I can add up $BD$ and $DC$ to get $8 = 3k + 5k.$
Solving for $k,$ I get $k = 1.$
So, $BD = 3$ and $DC = 5.$
Now, I can use the Stewart's theorem, which relates the lengths of the sides and the angle bisector in a triangle.
Stewart's theorem says that $AB^2 \cdot DC + AC^2 \cdot BD = AD^2 \cdot BC + BD \cdot DC \cdot BC.$
Plugging in the values, I get $6^2 \cdot 5 + 10^2 \cdot 3 = AD^2 \cdot 8 + 3 \cdot 5 \cdot 8.$
Simplifying, I get $180 + 300 = AD^2 \cdot 8 + 120.$
Subtracting 120 from both sides, I get $360 = AD^2 \cdot 8.$
Dividing by 8, I get $AD^2 = 45.$