2004 AMC 10B Problem 24

Question

In

\triangle ABC

we have

AB = 7,

AC = 8,

and

BC = 9.

Point

D

is on the circumscribed circle of the triangle so that

\overline{AD}

bisects

\angle BAC.

What is the value of

\frac{AD}{CD}?

\dfrac{9}{8}

\dfrac{5}{3}

2

\dfrac{17}{7}

\dfrac{5}{2}

Let

\overline{AD}

meet

\overline{BC}

at

E;

inscribed angles give

\angle ABE = \angle ADC

Triangles

ABE

and

ADC

are similar, so

\frac{AD}{CD} = \frac{AB}{BE};

find

BE

from the Angle Bisector Theorem

Let

\overline{AD}

meet

\overline{BC}

at

E.

Since

\angle ABC

and

\angle ADC

subtend the same arc, they are equal, and

\angle EAB = \angle CAD,

so

\triangle ABE \sim \triangle ADC.

Hence

\dfrac{AD}{CD} = \dfrac{AB}{BE}.

By the Angle Bisector Theorem,

\dfrac{BE}{EC} = \dfrac{AB}{AC},

so

BE = \dfrac{AB \cdot BC}{AB + AC} = \dfrac{7 \cdot 9}{15}.

Therefore

\begin{aligned} \dfrac{AD}{CD} &= \dfrac{AB}{BE} = \dfrac{AB + AC}{BC} \\ &= \dfrac{15}{9} = \dfrac{5}{3}. \end{aligned}

Thus, the correct answer is B.

Problem 24 in other years