Angle bisectors

Note that since $BY$ is both an angle bisector and a median, triangle $ABC$ is isosceles with $AB=BC$. Therefore, $BC = \boxed{16}$. A simple application of the angle bisector theorem (using $CZ$ as the angle bisector) yields $\frac{BZ}{AZ} = \frac{BC}{AC} = \frac{1}{2},$ so $BZ = \frac{1}{3} \cdot 16 = \boxed{ \frac{16}{3}}.$