In triangle $ABC$, $\angle ABC = 90^\circ$, and point $D$ lies on segment $BC$ such that $AD$ is an angle bisector.  If $AB = 105$ and $BD = 42$, then find $AC$.
By the angle bisector theorem, $AC/CD = AB/BD = 105/42 = 5/2$.  Let $AC = 5x$ and $CD = 2x$.

[asy]
unitsize(0.03 cm);

pair A, B, C, D;

A = (0,105);
B = (0,0);
C = (100,0);
D = (42,0);

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("$42$", (B + D)/2, S);
label("$105$", (A + B)/2, W);
label("$2x$", (C + D)/2, S);
label("$5x$", (A + C)/2, NE);
[/asy]

Then by Pythagoras, $(2x + 42)^2 + 105^2 = (5x)^2$.  This simplifies to $21x^2 - 168x - 12789 = 0$, which factors as $21(x - 29)(x + 21) = 0$, so $x = 29$.  Therefore, $AC = 5x = \boxed{145}$.