### Instruction: 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 $\cos \angle ABM$?### Response: I want to find $\cos \angle ABM$, which is $\frac{AM}{AB}$ by the definition of cosine.
To find $AM$, I need to know the length of $AC$, which I can get using the Pythagorean theorem.
$AC = \sqrt{AB^2 + BC^2} = \sqrt{10^2 + 24^2} = \sqrt{676} = 26$.
Now, how can I use the fact that $BM$ is an angle bisector?
Well, one property of angle bisectors is that they divide the opposite side in the same ratio as the adjacent sides.
That means $\frac{AM}{MC} = \frac{AB}{BC}$.
Plugging in the known values, I get $\frac{AM}{26 - AM} = \frac{10}{24}$.
Cross-multiplying, I get $24AM = 10(26 - AM)$.
Simplifying, I get $34AM = 260$, so $AM = \frac{260}{34} = \frac{130}{17}$.
Therefore, $\cos \angle ABM = \frac{AM}{AB} = \frac{130}{17} \div 10 = \frac{13}{17}$.