jmc

We might try sketching a diagram: Since $BC$ is a diameter of the circle, that makes $\angle BDC$ a right angle. Then, by $AA$ similarity, we see that $△ADB\sim △BDC\sim △ABC.$ Then, $\frac{BD}{AD}=\frac{CD}{BD},$ so $CD=\frac{B{D}^2}{AD}=\frac{4^2}{1}=\boxed{16}.$