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 a median of $\triangle ABC$, then what is $\cos \angle ABM$?
[asy]
pair A,B,C,M;
B = (0,0);
A = (0,10);
C = (24,0);
M = (A+C)/2;
draw(M--B--A--C--B);
label("$B$",B,SW);
label("$A$",A,N);
label("$C$",C,SE);
label("$M$",M,NE);
draw(rightanglemark(C,B,A,30));
[/asy]

The Pythagorean Theorem gives us $AC = \sqrt{AB^2 + BC^2} = \sqrt{100+576} = \sqrt{676}=26$.

The median to the hypotenuse of a right triangle has length half the hypotenuse, so $BM = AM$, which means $\angle ABM = \angle BAM$.  Therefore, we have $\cos \angle ABM = \cos \angle BAM = \cos\angle BAC = \frac{AB}{AC} = \frac{10}{26} = \boxed{\frac{5}{13}}$.