2008 AMC 10B Problem 6

Both conditions compare a piece with the whole segment, so pick a handy length for $AD$ and find where $B$ and $C$ land. A length of $10$ works well, since $10$ splits evenly into $5$ parts and into $10$ parts.

1. Let $AD = 10$.
2. $AB$ is $4$ times $BD$. So if $BD$ is $1$ part, $AB$ is $4$ parts, and $AD$ is $4 + 1 = 5$ parts. Each part is $10 \div 5 = 2$, so $BD = 2$.
3. $AC$ is $9$ times $CD$. So $AD$ is $9 + 1 = 10$ parts, each of length $1$, and $CD = 1$.
4. Both points are measured back from $D$: $B$ is $2$ away from $D$ and $C$ is $1$ away from $D$. So $C$ lies between $B$ and $D$, and
$$BC = BD - CD = 2 - 1 = 1.$$
5. As a fraction of $AD$: $\dfrac{BC}{AD} = \dfrac{1}{10}$.

The answer is $\boxed{\textbf{(C)}\ 1/10}$.

The phrase "$AB$ is $4$ times $BD$" cuts the whole segment into $4 + 1 = 5$ equal parts. Turning each ratio into a fraction of the same whole, $AD$, puts both points on one ruler, and then $BC$ is just a subtraction. Choosing a length that both $5$ and $10$ divide keeps the numbers whole.