jmc

Without loss of generality, suppose that $BA<BC$. Since $BD$ is the angle bisector of $\angle B$, by the Angle Bisector Theorem, it follows that $$\frac{AD}{CD}=\frac{BA}{BC}=\frac{3}{4}.$$ Thus, $AD<CD$, so $CD$ is the longer subsegment of $AC$. Solving for $AD$, it follows that $AD=\frac{3CD}{4}$. Also, we know that $AD+CD=AC=10$, and substituting our previous value for $AD$, we find that $\frac{3CD}{4}+CD=\frac{7CD}{4}=10⟹CD=\boxed{\frac{40}{7}}$ inches.