Check the picture below.
the idea being, that if we run an altitude segment from a right-angle in a triangle, and parallel to the opposite side, like in thise case, we end up with 3 similar triangles, a Large one, containing the other smaller ones, a Medium and a Small.
now, let's simply use proportions for those similar triangles.
[tex]\bf \stackrel{Large}{\cfrac{12}{4+x}}=\stackrel{Small}{\cfrac{4}{12}}\implies \cfrac{12}{4+x}=\cfrac{1}{3}\implies 36=4+x\implies \boxed{32=x} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{Small}{\cfrac{4}{y}}=\stackrel{Medium}{\cfrac{y}{x}}\implies 4x=y^2\implies 4(32)=y^2\implies 128=y^2\implies \sqrt{128}=y \\\\\\ \sqrt{64\cdot 2}=y\implies \sqrt{8^2\cdot 2}=y\implies \boxed{8\sqrt{2}=y} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \stackrel{Large}{\cfrac{z}{4+x}}=\stackrel{Medium}{\cfrac{x}{z}}\implies z^2=(4+x)x\implies z^2=4x+x^2\implies z = \sqrt{4x+x^2} \\\\\\ z = \sqrt{4(32)+32^2}\implies z = \sqrt{128+1024}\implies z = \sqrt{1152} \\\\\\ z = \sqrt{576\cdot 2}\implies z = \sqrt{24^2\cdot 2}\implies \boxed{z = 24\sqrt{2}}[/tex]
