smc

Let $△ABC$ be a right triangle with right angle at $B$, $AB=2$, and $BC=1$. Let the altitude from $B$ to $AC$ intersect $AC$ at $D$. From the Pythagorean Theorem, $AC=\sqrt{1^2+2^2}=\sqrt{5}$, so $BD=\frac{1\cdot 2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$. Since $△ADB\sim △BDC\sim △ABC$, $\frac{AD}{BD}=2$ and $\frac{CD}{BD}=\frac{1}{2}$. Therefore, $AD=\frac{4\sqrt{5}}{5}$ and $CD=\frac{\sqrt{5}}{5}$, so $CD:AD=\boxed{1:4}$.