If the ratio of the legs of a right triangle is $1: 2$, then the ratio of the corresponding segments of 
the hypotenuse made by a perpendicular upon it from the vertex is: 
$\textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5$
Let $\triangle ABC$ 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, $AC = \sqrt{1^2 + 2^2} = \sqrt5$, so $BD = \frac{1\cdot 2}{\sqrt5} = \frac{2\sqrt5}{5}$. Since $\triangle ADB \sim \triangle BDC \sim \triangle ABC$, $\frac{AD}{BD} = 2$ and $\frac{CD}{BD} = \frac12$. Therefore, $AD = \frac{4\sqrt5}{5}$ and $CD = \frac{\sqrt5}{5}$, so $CD:AD = \boxed{\textbf{(A) }1:4}$.