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 $\cos \angle ABM$?
707,106,781
cos∠ABM = cos∠(45º) = --------------------
1,000,000,000
+501 
