In $\triangle ABC$, $AB = 6$, $AC = 4$, $BC = 10$. $AD$ bisects $\angle A$, meets $BC$ at $D$. Then $BD = ?$

By the Internal Bisector Theorem, $BD:DC = AB:AC = 6:4 = 3:2$. So $BD = \dfrac{3}{3+2} \times BC = \dfrac{3}{5} \times 10 = 6$.