2004 AMC 10B Problem 24

Because $AD$ bisects $\angle BAC$, the inscribed angles $\angle BAD$ and $\angle DAC$ are equal. Equal inscribed angles intercept equal arcs, so the chords $BD$ and $CD$ are equal; call their common length $x$.

The points $A, B, D, C$ lie on one circle in that order, so $ABDC$ is a cyclic quadrilateral with diagonals $AD$ and $BC$. Ptolemy's theorem says the product of the diagonals equals the sum of the products of opposite sides:
$$AD \cdot BC = AB \cdot CD + AC \cdot BD.$$
Substituting the known lengths,
$$9 \cdot AD = 7x + 8x = 15x \quad\Longrightarrow\quad \frac{AD}{x} = \frac{15}{9} = \frac{5}{3}.$$

Since $x = CD$, the requested ratio is $\frac{AD}{CD} = \frac{5}{3}$.

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

The second intersection of an angle bisector with the circumcircle is the midpoint of the opposite arc, so it is equidistant from the other two vertices; that symmetry is what makes the configuration tractable. With a cyclic quadrilateral and all four sides plus one diagonal in play, Ptolemy's theorem is the natural tool, and it delivers the ratio without ever computing an actual length.

Let $E = AD \cap BC$. Angles $\angle ABE$ and $\angle ADC$ both stand on arc $AC$, and $\angle BAE = \angle DAC$, so $\triangle ABE \sim \triangle ADC$ and $\frac{AD}{CD} = \frac{AB}{BE}$. The angle bisector theorem gives $BE : EC = 7 : 8$, so $BE = \frac{7}{15} \cdot 9 = \frac{21}{5}$, and $\frac{AD}{CD} = \frac{7}{21/5} = \frac{5}{3}$.

• Not noticing that BD = CD (the bisected angle subtends equal arcs), which is the fact that collapses Ptolemy's relation into a single ratio.
• Confusing $D$ (on the circumcircle) with the foot of the bisector on $BC$, and reporting a ratio from the angle bisector theorem such as $\frac{9}{8}$ or $\frac{17}{7}$.