Triangle $ABC$ has a right angle at $B$, $AB=1$, and $BC=4$. The angle bisector of $\angle A$ intersects side $\overline{BC}$ at $D$. What is $BD$?

$AC=\sqrt{1^2+4^2}=\sqrt{17}$ by the Pythagorean theorem.
By the angle bisector theorem (theorem 8), $AB:AC=BD:DC$, so
$$BD=\frac{AB}{AB+AC}\cdot BC=\frac{1}{1+\sqrt{17}}\cdot 4=\frac4{1+\sqrt{17}}=\boxed{\frac{\sqrt{17}-1}4}.$$