ΔADC and ΔCDB are similar. Therefore the sides are in proportion:
[tex]\dfrac{DC}{DA}=\dfrac{DB}{DC}[/tex]
Substitute:
[tex]\dfrac{y}{4}=\dfrac{12}{y}[/tex] cross multiply
[tex](y)(y)=(4)(12)\\\\y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\cdot\sqrt3\\\\\boxed{y=4\sqrt3}[/tex]
