The sides of triangle $BAC$ are in the ratio $2:3:4$. $BD$ is the angle-bisector
drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$.
If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is:
$\text{(A)} \ 3\frac{1}{2} \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac{5}{7} \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac{1}{2}$
By the [Angle Bisector Theorem](https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem), we have $\frac{BA}{AD}=\frac{BC}{CD}$ which implies $\frac{AD}{DC}=\frac{BA}{BC}=\frac{3}{4}$. So $AC=10=AD+DC=\frac{7}{4}DC$, and thus $DC=\frac{40}{7}=5\frac{5}{7}$.
Hence
$\fbox{C}$