2002 AMC 12A Problem 23

Question

In triangle

ABC,

side

\overline{AC}

and the perpendicular bisector of

\overline{BC}

meet in point

D,

and

\overline{BD}

bisects

\angle ABC.

If

AD = 9

and

DC = 7,

what is the area of triangle

ABD?

14

21

28

14\sqrt{5}

28\sqrt{5}

D

is on the perpendicular bisector of

\overline{BC},

so

DB = DC = 7

The angle bisector gives

\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7};

apply the Law of Cosines with

\cos\theta

read from

\triangle BDC

Since

D

lies on the perpendicular bisector of

\overline{BC},

DB = DC = 7.

The angle bisector

\overline{BD}

gives

\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7},

so write

AB = 9x

and

BC = 7x.

Let

\theta = \angle ABD = \angle DBC.

In isosceles

\triangle BDC,

the foot of the perpendicular is the midpoint

M

of

\overline{BC},

so

\cos\theta = \dfrac{BM}{BD} = \dfrac{\frac{7x}{2}}{7} = \dfrac{x}{2}.

Applying the Law of Cosines in

\triangle ABD:

9^2 = (9x)^2 + 7^2 - 2(9x)(7)\cdot\dfrac{x}{2},

which simplifies to

81 = 18x^2 + 49,

so

x = \dfrac43

and

AB = 12.

Now

\triangle ABD

has sides

9, 7, 12.

By Heron’s formula with

s = 14,

the area is

\sqrt{14\cdot 5\cdot 7\cdot 2} = \sqrt{980} = 14\sqrt5.

Thus, the correct answer is D.

Problem 23 in other years