In triangle $ABC$, $\angle ABC = 90^\circ$, and point $D$ lies on segment $\overline{BC}$ such that $\overline{AD}$ is an angle bisector.  If $AB = 12$ and $BD = 4$, then find $AC$.
By the angle bisector theorem, $AC/CD = AB/BD = 12/4 = 3$.  Let $AC = 3x$ and $CD = x$.

[asy]
unitsize(0.3 cm);

pair A, B, C, D;

A = (0,12);
B = (0,0);
C = (9,0);
D = (4,0);

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

label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$4$", (B + D)/2, S);
label("$12$", (A + B)/2, W);
label("$x$", (C + D)/2, S);
label("$3x$", (A + C)/2, NE);
[/asy]

Then by Pythagoras, $(x + 4)^2 + 12^2 = (3x)^2$.  This simplifies to $8x^2 - 8x - 160 = 0$, which factors as $8(x - 5)(x + 4) = 0$, so $x = 5$.  Therefore, $AC = 3x = \boxed{15}$.