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 $AD^2.$
First of all, a sketch might be useful: [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 6);
pB = (0, 0);
pC = (-8, 0);
pD = (pB * 10 + pC * 6) / (14);
draw(pA--pB--pC--pA);
draw(pA--pD);
label("$A$", pA, N);
label("$B$", pB, SE);
label("$C$", pC, SW);
label("$D$", pD, S);
[/asy] It is clear that we have a $3:4:5$ right triangle on our hands, so we have a right angle at $B.$ That means if we can find $DB,$ we can use the Pythagorean Theorem on $\triangle ABD$ to find $AD^2.$

To find $DB,$ we see that $DB:DC = AB:AC = 6:10$ thanks to the Angle Bisector Theorem. That means $DB:DC:BC = 6:10:16,$ since $BC = DB + DC.$ So $DB = \frac{6}{16} \cdot BC = 3.$

Finally, we have that $AD^2 = AB^2 + BD^2 = 6^2 + 3^2 = 36 + 9 = \boxed{45}.$