Look at the picture.
[tex]\Delta ABD\ and\ \Delta BCD\ are\ similar,\ therefore:\\\\\dfrac{y}{4}=\dfrac{16}{y}\ \ \ \ |cross\ multiply\\\\y\cdot y=4\cdot16\\\\y^2=64\to y=\sqrt{64}\to y=8[/tex]
Use Pythagorean theory:
[tex]z^2=4^2+8^2\\\\z^2=16+64\\\\z^2=80\to z=\sqrt{80}\\\\z=\sqrt{16\cdot5}\\\\z=\sqrt{16}\cdot\sqrt5\\\\\boxed{z=4\sqrt5}[/tex]
