In right triangle $ABC$, we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$. If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is an angle bisector of $\triangle ABC$, then what is $\cos \angle ABM$?
Since $\angle ABC = 90^\circ$ and $\overline{BM}$ bisects $\angle ABC$, we have $\angle ABM = 45^\circ$, so $\cos \angle ABM = \cos 45^\circ = \boxed{\frac{\sqrt{2}}{2}}$.

[asy]
unitsize(0.25 cm);

pair A, B, C, M;

A = (0,10);
B = (0,0);
C = (24,0);
M = extension(B, B + dir(45), A, C);

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

label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$M$", M, NE);
[/asy]