2009 AMC 12B Problem 16

Attempt Problem 16 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Trapezoid

ABCD

has

AD \parallel BC,

BD = 1,

\angle DBA = 23^\circ,

and

\angle BDC = 46^\circ.

The ratio

BC : AD

is

9 : 5.

What is

CD?

\dfrac{7}{9}

\dfrac{4}{5}

\dfrac{13}{15}

\dfrac{8}{9}

\dfrac{14}{15}

Answer: B

Concepts:angle bisector theorem trapezoid parallel lines

Difficulty rating: 1800

Small hint:

Draw the line through

D

parallel to

AB,

meeting

BC

at

E

; then

BE = AD

Big hint:

Show

DE

bisects

\angle BDC,

then apply the angle bisector theorem in

\triangle BDC

Solution:

Draw the line through

D

parallel to

AB,

meeting

BC

at

E,

so

ABED

is a parallelogram with

BE = AD.

Then

\angle BDE = \angle DBA = 23^\circ,

and since

\angle BDC = 46^\circ,

segment

DE

bisects

\angle BDC.

By the angle bisector theorem in

\triangle BDC,

\dfrac{EC}{BE} = \dfrac{DC}{DB},

so

\begin{aligned} CD &= DB \cdot \dfrac{BC - AD}{AD} \\ &= 1 \cdot \left(\dfrac{9}{5} - 1\right) \\ &= \dfrac{4}{5}. \end{aligned}

Thus, the correct answer is B.

Problem 16 in other years