2009 AMC 10B Problem 20

The angle bisector theorem is the standard tool here: it turns "this line cuts the angle in half" into a statement about lengths.

1. First find the third side. By the Pythagorean theorem, $AC = \sqrt{1^2 + 2^2} = \sqrt5$.
2. The angle bisector theorem says: the bisector from $A$ splits side $BC$ into two pieces in the same ratio as the two sides that meet at $A$. So
$$BD : DC = AB : AC = 1 : \sqrt5.$$
3. That means $BC$ is cut into $1 + \sqrt5$ equal "shares," and $BD$ gets $1$ of them. Since $BC = 2$,
$$BD = \frac{2}{1 + \sqrt5}.$$
4. The answers have no square root on the bottom, so multiply top and bottom by $\sqrt5 - 1$. On the bottom, $(\sqrt5 + 1)(\sqrt5 - 1) = 5 - 1 = 4$:
$$BD = \frac{2(\sqrt5 - 1)}{4} = \frac{\sqrt5 - 1}{2}.$$
5. Sense check: $\sqrt5 \approx 2.24$, so $BD \approx 0.62$. That is less than half of $BC$, which makes sense because $AB$ is the shorter of the two sides at $A$.

The answer is $\boxed{\textbf{(B)}\ \frac{\sqrt5 - 1}{2}}$.

The angle bisector theorem turns an angle fact into a ratio of lengths. The ratio uses the two sides that form the bisected angle ($AB$ and $AC$), not the side being cut. Once you have the ratio and the total length, splitting the side is one line of arithmetic.

Drop a perpendicular from $D$ to $AC$ and call its foot $F$. Every point on the bisector is the same distance from both sides of the angle, so $DF = DB$. Call this length $d$. Triangles $ABD$ and $AFD$ are right triangles with the same slanted side $AD$ and the same leg $d$, so $AF = AB = 1$, and $FC = \sqrt5 - 1$.

Now triangle $DFC$ has a right angle at $F$ and shares angle $C$ with triangle $ABC$, so they have the same shape. In triangle $ABC$ the side across from $C$ is $1$ and the side next to $C$ is $2$, so in triangle $DFC$ the side across from $C$ is half the side next to it:
$$d = \tfrac12 \times FC = \frac{\sqrt5 - 1}{2}.$$