2009 AMC 10B Problem 20

By the Pythagorean Theorem, $AC=\sqrt{1^2+2^2}=\sqrt5.$ The Angle Bisector Theorem gives $$\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5},$$ so $DC=\sqrt5\,BD.$ Since $BD+DC=2,$ we have $BD(1+\sqrt5)=2,$ so $$BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}.$$ Thus, the correct answer is B.