2004 AMC 10B Problem 24

1. Equal angles give equal chords. $AD$ splits angle $BAC$ into two equal angles, $\angle BAD$ and $\angle DAC$. Both have their corner on the circle. An angle with its corner on the circle is half the arc it cuts off, so equal angles cut off equal arcs: arc $BD$ equals arc $DC$. Equal arcs have equal chords, so $BD = CD$. Call this length $x$.
2. Look at the four points. $A$, $B$, $D$, $C$ all lie on the circle, in that order. They form a four-sided figure $ABDC$ whose diagonals are $AD$ and $BC$.
3. Use Ptolemy's theorem. For four points in order on a circle, it says: (one diagonal) $\times$ (other diagonal) $=$ (sum of the products of opposite sides). Here the opposite sides are $AB$ with $DC$, and $AC$ with $BD$:
$$AD \cdot BC = AB \cdot CD + AC \cdot BD.$$
4. Put in the numbers. $BC = 9$, $AB = 7$, $AC = 8$, and $CD = BD = x$:
$$9 \cdot AD = 7x + 8x = 15x.$$
5. So $\frac{AD}{x} = \frac{15}{9} = \frac{5}{3}$. Since $x = CD$, this is the ratio we want.

The answer is $\boxed{\textbf{(B)}\ \frac{5}{3}}$.

When an angle bisector from a corner is extended to hit the circle again, it lands halfway around the opposite arc, so that point is the same distance from the other two corners. With four points on a circle and several side lengths known, Ptolemy's theorem is the natural tool, and here it gives the ratio without finding any actual length.

Use similar triangles if you do not know Ptolemy's theorem. Let $E$ be where $AD$ crosses $BC$.

1. $\angle ABC$ and $\angle ADC$ both cut off arc $AC$, so they are equal. Also $\angle BAE = \angle DAC$ (the bisected halves). So triangles $ABE$ and $ADC$ have two equal angles and are similar. Matching sides gives $\frac{AD}{CD} = \frac{AB}{BE}$.
2. The angle bisector theorem says $AE$ cuts $BC$ in the ratio $AB : AC = 7 : 8$. So $BE = \frac{7}{15} \times 9 = \frac{21}{5}$.
3. Then $\frac{AD}{CD} = 7 \div \frac{21}{5} = \frac{35}{21} = \frac{5}{3}$.